VERY IMPORTANT WORD PROBLEMS (QUADRATIC EQUATION )

         VERY IMPORTANT WORD PROBLEMS                       (QUADRATIC EQUATION )

Definition of Quadratic Equation

An equation of second-degree polynomial in one variable, such as x$x$ usually equated to zero, is called a quadratic equation. The coefficient of x2$x^2$ must not be zero in a quadratic equation. If p(x)$p(x)$ is a quadratic polynomial, then p(x)=0$p(x)=0$ is called a quadratic equation.

Some examples are,

3x2+2x+2=0,−x2+6x+1=0,7x2−6x+4=0

Quadratic Equation Using Completing Square                    Method has  deleted in term 2 2022                    

the quadratic formula 
x=−b±b2−4ac√2a

                                                            

Q.1. A boat has a speed of 16 km/hr$16\mathrm{k}\mathrm{m}/\mathrm{h}\mathrm{r}$ in still water. In a particular river, it takes 2$2$ hours more to go 32 km$32\mathrm{k}\mathrm{m}$ upstream than to return to the same point. What is the speed at which water is flowing in the river?

Ans: Let the speed of the stream be xkm/ph$\mathrm{x }\text{km/ph}$.
We know that when the boat travels upstream, the relative speed between the boat and the stream (16−x)kmph$(16-x)\text{kmph}$
When the boat travels downstream, the relative speed between the boat and the stream (16+x)kmph$(16+x)\text{kmph}$
The time required to go upstream = Distance  Speed =3216−x$=\frac{\text{ Distance }}{\text{ Speed }}=\frac{32}{16-x}$ hours$\text{hours}$
Similarly, the time required to go downstream = Distance  Speed =3216+x$=\frac{\text{ Distance }}{\text{ Speed }}=\frac{32}{16+x}$ hours$\text{hours}$
Now, according to the question,
3216−x−3216+x=2⇒32[116−x−116+x]=2⇒[116−x−116+x]=232$\frac{32}{16-x}-\frac{32}{16+x}=2\Rightarrow 32[\frac{1}{16-x}-\frac{1}{16+x}]=2\Rightarrow [\frac{1}{16-x}-\frac{1}{16+x}]=\frac{2}{32}$
To solve the for x$x$ we need to convert the above equation into the standard form of the quadratic equation.
⇒[(16+x)−(16−x)(16−x)(16+x)]=116⇒[16+x−16+x(16−x)(16+x)]=116$\Rightarrow \left[\frac{(16+x)-(16-x)}{(16-x)(16+x)}\right]=\frac{1}{16}\Rightarrow \left[\frac{16+x-16+x}{(16-x)(16+x)}\right]=\frac{1}{16}$
⇒[2x162−x2]=116$\Rightarrow \left[\frac{2x}{{16}^2-x^2}\right]=\frac{1}{16}$ (using the algebraic identity a2−b2=(a+b)(a−b)$a^2-b^2=(a+b)(a-b)$
Now, cross multiplying we have, ⇒32x=256−x2$\Rightarrow 32x=256-x^2$
We have ⇒x2+32x−256=0$\Rightarrow +32x-256=0$
⇒x2+2×16×x−256=0$\Rightarrow x^2+2\times 16\times x-256=0$
Now, to make the perfect square, we need to add and subtract (16)2$(16{)}^2$ from LHS ⇒x2+2×16×x+(16)2−(16)2−256=0⇒(x+16)2−(16)2−256=0$\Rightarrow x^2+2\times 16\times x+(16{)}^2-(16{)}^2-256=0\Rightarrow (x+16{)}^2-(16{)}^2-256=0$
Transferring −(16)2−256$-(16{)}^2-256$ from LHS to RHS we have,
⇒(x+16)2=512$\Rightarrow (x+16{)}^2=512$
Taking square root on both sides we have, ⇒x+16=±512−−−√$\Rightarrow x+16=\pm \sqrt{512}$
⇒x=±162–√−16$\Rightarrow x=\pm 16\sqrt{2}-16$
⇒x=+162–√−16$\Rightarrow x=+16\sqrt{2}-16$ or, x=−162–√−16$x=-16\sqrt{2}-16$
We will consider the positive value of x$x$ as speed can not be negative. Hence, the speed at which the water is flowing in the river is +162–√–16km/ph

Q.2. Priya decides to build a prayer hall of her house having a carpet area of 300 squaremeters with its length of 25 metres more than its breadth. What should be the breadth and length of the prayer hall?

Ans: Let us assume breadth is x$x$, and the length is (x+25)$(x+25)$. According to the question,
x(x+25)=300⇒x2+25x−300=0$x(x+25)=300\Rightarrow x^2+25x-300=0$
⇒$\Rightarrow$ From the given quadratic equation a=1,b=25,c=−300$a=1,b=25,c=-300$
Quadratic equation formula is given by x=−b±b2−4ac√2a$x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$
x=−(25)±(25)2−4×1×(−300)√2×1=−25±1825√2$x=\frac{-(25)\pm \sqrt{(25{)}^2-4\times 1\times (-300)}}{2\times 1}=\frac{-25\pm \sqrt{1825}}{2}$
x=−25+1825√2=8.86$x=\frac{-25+\sqrt{1825}}{2}=8.86$ (approx) or x=−25−1825√2=−33.86$x=\frac{-25-\sqrt{1825}}{2}=-33.86$ (approx) We will avoid the negative value of x$x$ as breadth can not be negative.
Hence, the breadth is, x=8.86$x=8.86$ (approx) and the length is x+25=8.86$x+25=8.86$ (approx) +25=33.86$+25=33.86$ (approx)

Q.3. The product of two odd consecutive positive integers is 63$63$. Find the numbers.

Ans: Let the two odd integers be x,x+2$x,x+2$. We have:
Now, according to the statement,
x(x+2)=63⇒x2+2x–63=0$x\left(x+2\right)=63\Rightarrow x^2+2x–63=0$
⇒x2+(9–7)x–63=0⇒x2+9x–7x–63=0$\Rightarrow x^2+\left(9–7\right)x–63=0\Rightarrow x^2+9x–7x–63=0$
⇒x(x+9)–7(x+9)=0⇒(x+9)(x–7)=0$\Rightarrow x\left(x+9\right)–7\left(x+9\right)=0\Rightarrow \left(x+9\right)\left(x–7\right)=0$
So, x=−9$x=-9$ or, x=7$x=7$
Here, we need to find the positive odd integers. So, avoid the negative value of x$x$.
Hence, the two odd consecutive integers are x=7$x=7$ and x+2=7+2=9$x+2=7+2=9$.

Q.4. The product of Rime’s age two years ago and her age four years from now is one more than twice her present age. What is her present age?

Ans: Let her present age be x$x$.
Two years ago, her age was (x−2)$(x-2)$.
Her age after four years from now is (x+4)$(x+4)$.
One more than twice her present age is (1+2x)$(1+2x)$.
According to the statement,
(x−2)(x+4)=1+2xx2+4x−2x−8=1+2x$(x-2)(x+4)=1+2x{x}^2+4x-2x-8=1+2x$
x2+2x−2x−8−1=0$x^2+2x-2x-8-1=0$
x2−9=0x2=9$x^2-9=0x^2=9$
Thus, x=±3$x=\pm 3$
Now, age can not be negative.
Hence, Rime’s present age is 3$3$ years

Q.5. If the speed of a car is increased by 10km/hr the time of journey for a distance of 72km is reduced by 36minutes Find the initial speed of the car.

Ans: Let the initial speed of the car is xkm/ph.
Speed of the car after increasing the speed is (x+10),kmph$(x+10),\text{kmph}$.
Distance is 72 km$72\mathrm{k}\mathrm{m}$ (given).
Time taken to cover the distance is (72x)hours$\left(\frac{72}{x}\right)\text{hours}$
Time taken after increasing the speed is (72x+10)hours$\left(\frac{72}{x+10}\right)\text{hours}$
According to the question,
⇒(1x)–(1x+10)=3660×72$\Rightarrow \left(\frac{1}{x}\right)–\left(\frac{1}{x+10}\right)=\frac{36}{60\times 72}$
⇒[x+10–xx(x+10)]=1120⇒x2+10x–1200=0$\Rightarrow \left[\frac{x+10–x}{x\left(x+10\right)}\right]=\frac{1}{120}\Rightarrow x^2+10x–1200=0$
⇒x2+40x–30x–1200=0$\Rightarrow x^2+40x–30x–1200=0$
⇒x(x+40)–30(x+40)=0$\Rightarrow x\left(x+40\right)–30\left(x+40\right)=0$
⇒(x+40)(x–30)=0⇒x=–40$\Rightarrow \left(x+40\right)\left(x–30\right)=0\Rightarrow x=–40$ or x=30x=30

Solve by factorization

a. 4x2−4a2x+(a4−b4)=0$4x^2-4a^{2}x+\left(a^4-b^4\right)=0$

Ans: 

4x2−[2(a2+b2)+2(a2−b2)] x+(a2−b2)(a2+b2)=0$4x^2-\left[2\left(a^2+b^2\right)+2\left(a^2-b^2\right)\right]x+\left(a^2-b^2\right)\left(a^2+b^2\right)=0$ 

⇒2x[2x−(a2+b2)]−(a2−b2)[2x−(a2+b2)]=0$\Rightarrow 2x\left[2x-\left(a^2+b^2\right)\right]-\left(a^2-b^2\right)\left[2x-\left(a^2+b^2\right)\right]=0$ 

⇒x=a2+b22,x=a2−b22$\Rightarrow x={\displaystyle \frac{a^2+b^2}{2}},x={\displaystyle \frac{a^2-b^2}{2}}$

Therefore, x=a2+b22(or)x=a2−b22$x={\displaystyle \frac{a^2+b^2}{2}}\left(or\right)x={\displaystyle \frac{a^2-b^2}{2}}$

b. x2+(aa+bx+a+ba)x+1=0$x^2+\left({\displaystyle \frac{a}{a+b}}x+{\displaystyle \frac{a+b}{a}}\right)x+1=0$

Ans:

            x2+(aa+bx+a+ba)x+1=0$x^2+\left({\displaystyle \frac{a}{a+b}}x+{\displaystyle \frac{a+b}{a}}\right)x+1=0$ 

            ⇒x2+(aa+bx+a+bax+aa+b.a+ba)=0$\Rightarrow x^2+\left({\displaystyle \frac{a}{a+b}}x+{\displaystyle \frac{a+b}{a}}x+{\displaystyle \frac{a}{a+b}}.{\displaystyle \frac{a+b}{a}}\right)=0$ 

            ⇒[x+aa+b]+a+ba[x+aa+b]=0$\Rightarrow \left[x+{\displaystyle \frac{a}{a+b}}\right]+{\displaystyle \frac{a+b}{a}}\left[x+{\displaystyle \frac{a}{a+b}}\right]=0$ 

            ⇒x=−aa+b,x=(−a+b)a,a+b=0$\Rightarrow x=-{\displaystyle \frac{a}{a+b}},x={\displaystyle \frac{\left(-a+b\right)}{a}},a+b=0$ 

c. 1a+b+x=1a+1b+1x,(a+b≠0)${\displaystyle \frac{1}{a+b+x}}={\displaystyle \frac{1}{a}}+{\displaystyle \frac{1}{b}}+{\displaystyle \frac{1}{x}},\left(a+b\ne 0\right)$

Ans:

1a+b+x=1a+1b+1x${\displaystyle \frac{1}{a+b+x}}={\displaystyle \frac{1}{a}}+{\displaystyle \frac{1}{b}}+{\displaystyle \frac{1}{x}}$ 

⇒1a+b+x−1x=1a+1b$\Rightarrow {\displaystyle \frac{1}{a+b+x}}-{\displaystyle \frac{1}{x}}={\displaystyle \frac{1}{a}}+{\displaystyle \frac{1}{b}}$ 

⇒x−(a+b+x)x(a+b+x)=a+bab$\Rightarrow {\displaystyle \frac{x-\left(a+b+x\right)}{x\left(a+b+x\right)}}={\displaystyle \frac{a+b}{ab}}$ 

⇒(a+b){x(a+b+x)+ab}=0$\Rightarrow \left(a+b\right)\left\{x\left(a+b+x\right)+ab\right\}=0$ 

⇒x(a+b+x)+ab=0$\Rightarrow x\left(a+b+x\right)+ab=0$ 

⇒x2+ax+bx+ab=0$\Rightarrow x^2+ax+bx+ab=0$ 

⇒(x+a)(x+b)=0$\Rightarrow \left(x+a\right)\left(x+b\right)=0$ 

⇒x=−a(or)x=−b$\Rightarrow x=-a\left(or\right)x=-b$ 

d. (x−3)(x−4)34332$\left(x-3\right)\left(x-4\right){\displaystyle \frac{34}{{33}^2}}$

(x−3)(x−4)=34332$\left(x-3\right)\left(x-4\right)={\displaystyle \frac{34}{{33}^2}}$ 

⇒x2−7x+12=34332$\Rightarrow x^2-7x+12={\displaystyle \frac{34}{{33}^2}}$ 

⇒x2−7x+13034332=0$\Rightarrow x^2-7x+{\displaystyle \frac{13034}{{33}^2}}=0$ 

Ans:

⇒x2−7x+9833×13333=0$\Rightarrow x^2-7x+{\displaystyle \frac{98}{33}}\times {\displaystyle \frac{133}{33}}=0$ 

⇒x2−23133x+9833×13333=0$\Rightarrow x^2-{\displaystyle \frac{231}{33}}x+{\displaystyle \frac{98}{33}}\times {\displaystyle \frac{133}{33}}=0$ 

⇒x2x−23133x+9833×13333=0$\Rightarrow x^{2}x-{\displaystyle \frac{231}{33}}x+{\displaystyle \frac{98}{33}}\times {\displaystyle \frac{133}{33}}=0$ 

 ⇒x2−(9833+13333)x+9833×13333=0$\Rightarrow x^2-\left({\displaystyle \frac{98}{33}}+{\displaystyle \frac{133}{33}}\right)x+{\displaystyle \frac{98}{33}}\times {\displaystyle \frac{133}{33}}=0$ 

 ⇒x2−9833x−13333x+9833×13333=0$\Rightarrow x^2-{\displaystyle \frac{98}{33}}x-{\displaystyle \frac{133}{33}}x+{\displaystyle \frac{98}{33}}\times {\displaystyle \frac{133}{33}}=0$ 

 ⇒(x−9833)x−13333(x−9833)=0$\Rightarrow \left(x-{\displaystyle \frac{98}{33}}\right)x-{\displaystyle \frac{133}{33}}\left(x-{\displaystyle \frac{98}{33}}\right)=0$ 

 ⇒(x−9833)(x−13333)=0$\Rightarrow \left(x-{\displaystyle \frac{98}{33}}\right)\left(x-{\displaystyle \frac{133}{33}}\right)=0$ 

 ⇒x=9833(or)x=13333$\Rightarrow x={\displaystyle \frac{98}{33}}\left(or\right)x={\displaystyle \frac{133}{33}}$ 

e. x=12−12−12−xx ≠ 2$\text{x=}{\displaystyle \frac{1}{2-{\displaystyle \frac{1}{2-{\displaystyle \frac{1}{2-x}}}}}}x\ne 2$

Ans:

x=12−12−12−xx ≠ 2 $\text{x=}{\displaystyle \frac{1}{2-{\displaystyle \frac{1}{2-{\displaystyle \frac{1}{2-x}}}}}}x\ne 2$ 

 ⇒ x=12−12−12−x$\Rightarrow \text{ x=}{\displaystyle \frac{1}{2-{\displaystyle \frac{1}{2-{\displaystyle \frac{1}{2-x}}}}}}$ 

 ⇒x=12−12−(2−x)4−2x−1$\Rightarrow \text{x=}{\displaystyle \frac{1}{2-{\displaystyle \frac{1}{2-{\displaystyle \frac{\left(2-x\right)}{4-2x-1}}}}}}$ 

 ⇒x=12−2−x3−2x$\Rightarrow \text{x=}{\displaystyle \frac{1}{2-{\displaystyle \frac{2-x}{3-2x}}}}$ 

 ⇒x=3−2x2(3−2x)−(2−x)$\Rightarrow x={\displaystyle \frac{3-2x}{2\left(3-2x\right)-\left(2-x\right)}}$ 

 ⇒x=3−2x4−3x$\Rightarrow x={\displaystyle \frac{3-2x}{4-3x}}$ 

 ⇒4x−3x2 =3−2x$\Rightarrow 4x-3x^2=3-2x$ 

 ⇒3x2−6x+3=0 $\Rightarrow 3x^2-6x+3=0$ 

 ⇒ (x−1)2 =0 $\Rightarrow {\left(x-1\right)}^2=0$ 

 ⇒x =1, 1.$\Rightarrow x=1,1.$ 

2. By the method of completion of squares show that the equation 4x2+3x+5=0$\mathbf{4}{\mathbf{x}}^{\mathbf{2}}+\mathbf{3}\mathbf{x}+\mathbf{5}=\mathbf{0}$ has no real roots.

Ans:

 4x2+3x+5=0$4x^2+3x+5=0$ 

 ⇒x2+34x+54=0$\Rightarrow x^2+{\displaystyle \frac{3}{4}}x+{\displaystyle \frac{5}{4}}=0$ 

 ⇒x2+34x+(38)2=−54+964$\Rightarrow x^2+{\displaystyle \frac{3}{4}}x+{\left({\displaystyle \frac{3}{8}}\right)}^2=-{\displaystyle \frac{5}{4}}+{\displaystyle \frac{9}{64}}$ 

 ⇒(x+38)2=−7164$\Rightarrow {\left(x+{\displaystyle \frac{3}{8}}\right)}^2=-{\displaystyle \frac{71}{64}}$ 

 ⇒x+38=−7164−−−−√$\Rightarrow x+{\displaystyle \frac{3}{8}}=\sqrt{-{\displaystyle \frac{71}{64}}}$ 

 Which is not a real number. Hence the equation has no real roots.

3. The sum of areas of two squares is 468m2. If the difference of their perimeters is 24cm, find the sides of the two squares. 

Ans:

Let, the side of the larger square be x. 

Let, the side of the smaller square be y. 

x2+y2=468$x^2+y^2=468$ 

Cond. II 4x-4y = 24 

 ⇒xy=6$\Rightarrow xy=6$ 

 ⇒x=6+y$\Rightarrow x=6+y$ 

 ⇒x2+y2=468$\Rightarrow x^2+y^2=468$ 

 ⇒(6+y)2+y2= 468$\Rightarrow {\left(6+y\right)}^2+y^2=468$ 

on solving we get 

y = 12 

⇒ x = (12+6) = 18 m 

∴ The length of the sides of the two squares are 18m and 12m. 

4. A dealer sells a toy for Rs.24 and gains as much percent as the cost price of the toy. Find the cost price of the toy.

Ans:

Let the C.P be x 

∴Gain = x% 

⇒Gain =xx100$\Rightarrow Gain=x{\displaystyle \frac{x}{100}}$

S.P = C.P +Gain 

SP = 24 

⇒x+x2100=24$\Rightarrow x+{\displaystyle \frac{x^2}{100}}=24$

On solving we get x = 20 or x = -120 (reject this as cost cannot be negative) 

∴ C.P of toy = Rs.20 

5. A fox and an eagle lived at the top of a cliff of height 6m, whose base was at a distance of 10m from a point A on the ground. The fox descends the cliff and went straight to point A. The eagle flew vertically up to height x meters and then flew in a straight line to a point A, the distance traveled by each being the same. 3 Find the value of x.

Ans: 

Distance traveled by the fox = distance traveled by the eagle 

            (6+x)2+(10)2$(6+x{)}^2+(10{)}^2$ = (16–x)2$(16–x{)}^2$  

            on solving we get x = 2.72m. 

6. A lotus is 2m above the water in a pond. Due to wind, the lotus slides on the side and only the stem completely submerges in the water at a distance of 10m from the original position. Find the depth of water in the pond. 

Ans: 

From,above figure,We can write as,

(x+2)2=x2+102${(x+2)}^2=x^2+{10}^2$ 

⇒x2+4x+4$\Rightarrow x^2+4x+4$ = x2+100$x^2+100$ 

⇒4x+4=100$\Rightarrow 4x+4=100$

⇒x=24$\Rightarrow x=24$

Therefore, the depth of water in the pond is 24m. 

7. Solve x=6+6+6.....−−−−√−−−−−−−−√−−−−−−−−−−−−−√$x=\sqrt{6+\sqrt{6+\sqrt{\mathrm{6.....}}}}$

Ans: 

 x=6+6+6.....−−−−√−−−−−−−−√−−−−−−−−−−−−−√$x=\sqrt{6+\sqrt{6+\sqrt{\mathrm{6.....}}}}$ 

 ⇒x=6+x−−−−−√$\Rightarrow x=\sqrt{6+x}$ 

 ⇒x2=6+x$\Rightarrow x^2=6+x$ 

 ⇒x2−x−6=0$\Rightarrow x^2-x-6=0$ 

 ⇒(x −3)(x + 2)=0$\Rightarrow \left(x-3\right)\left(x+2\right)=0$ 

 ⇒x = 3$\Rightarrow x=3$ 

As x cannot be negative x is not equal to 2.

8. The hypotenuse of a right triangle is 20m. If the difference between the length of the 4 other sides is 4m. Find the sides. 

Ans: 

(Image will be uploaded soon)

From above figure,

x2+y2=202$x^2+y^2={20}^2$ 

 x2+y2=400 $x^2+y^2=400$ 

 also x−y=4$alsox-y=4$ 

 ⇒x = 404 + y $\Rightarrow x=404+y$ 

 ⇒(4 + y)2+y2=400$\Rightarrow {\left(4+y\right)}^2+y^2=400$ 

 ⇒2y2+8y−384=0$\Rightarrow 2y^2+8y-384=0$ 

 ⇒(y + 16) (y  12)=0$\Rightarrow \left(y+16\right)\left(y12\right)=0$ 

 ⇒y = 12 (or)y=16(notpossible)$\Rightarrow y=12\left(or\right)y=16\left(notpossible\right)$ 

∴sides are 12cm and 16cm 

9. The positive value of k for which x2+ Kx + 64 = 0 & x2− 8x + k = 0${\mathbf{x}}^{\mathbf{2}}+\mathbf{K}\mathbf{x}+\mathbf{64}=\mathbf{0}\mathrm{\&}{\mathbf{x}}^{\mathbf{2}}-\mathbf{8}\mathbf{x}+\mathbf{k}=\mathbf{0}$will have real roots.

Ans:

x2+ Kx + 64 = 0⇒b2−4ac ≥ 0⇒K2−256 ≥ 0⇒K ≥ 16 or K ≤ − 16 …………… (1)x2−8x + K = 064  4K ≥ 0⇒4K ≤ 64⇒K ≤ 16 …………… (2)From (1) & (2) K = 16$\begin{array}{l}{x}^2+Kx+64=0\\ \Rightarrow b^2-4ac\ge 0\\ \Rightarrow K^2-256\ge 0\\ \Rightarrow K\ge 16orK\le -16\dots \dots \dots \dots \dots \left(1\right)\\ x^2-8x+K=0\\ 644K\ge 0\\ \Rightarrow 4K\le 64\\ \Rightarrow K\le 16\dots \dots \dots \dots \dots \left(2\right)\\ From\left(1\right)\mathrm{\&}\left(2\right)K=16\end{array}$

10. A teacher attempting to arrange the students for mass drill in the form of a solid square found that 24 students were left over. When he increased the size of the square by one student he found he was short of 25 students. Find the number of students. 

Ans: 

Let the side of the square be x.$x.$

No. of students = x2+ 24$x^2+24$

New side = x + 1$x+1$

No. of students = (x + 1)2 −25${\left(x+1\right)}^2-25$

⇒x2+ 24= (x + 1)2 −25⇒x2+ 24 = x2+ 2x +1 −25⇒2x = 48⇒x = 24$\begin{array}{l}\Rightarrow x^2+24={\left(x+1\right)}^2-25\\ \Rightarrow x^2+24=x^2+2x+1-25\\ \Rightarrow 2x=48\\ \Rightarrow x=24\end{array}$

∴ side of square = 24 

No. of students = 576 + 24 = 600$576+24=600$

11. A pole has to be erected at a point on the boundary of a circular park of diameter 13m in such a way that the differences of its distances from two diametrically opposite fixed gates A $ B on the boundary in 7m. Is it possible to do so? If the answer is yes at what distances from the two gates should the pole be erected? 

Ans: 

AB = 13 m 

BP = x 

(Image will be uploaded soon)

⇒AP − BP = 7⇒AP = x + 7 APQ⇒(13)2 = (x + 7)2+ x2⇒x2+7x  −60 = 0⇒(x + 12) (x  −5) = 0⇒x = − 12 (not possible) or x = 5$\begin{array}{l}\Rightarrow AP-BP=7\\ \Rightarrow AP=x+7APQ\\ \Rightarrow {\left(13\right)}^2={\left(x+7\right)}^2+x^2\\ \Rightarrow x^2+7x-60=0\\ \Rightarrow \left(x+12\right)\left(x-5\right)=0\\ \Rightarrow x=-12\left(notpossible\right)orx=5\end{array}$

∴Pole has to be erected at a distance of 5m from gate B & 12m from gate A. 

12. If the roots of the equation (a−b)x2+ b−c) x+ (c − a)= 0$\left(\mathbf{a}-\mathbf{b}\right){\mathbf{x}}^{\mathbf{2}}+\mathbf{b}-\mathbf{c})\mathbf{x}+\left(\mathbf{c}-\mathbf{a}\right)=\mathbf{0}$ are equal. Prove that2a=b + c$\mathbf{2}\mathbf{a}=\mathbf{b}+\mathbf{c}$. 

Ans: 

(ab)x2+(bc) x+(ca)=0Given:2a=b+cB2−4AC=0(bc)2[4(ab)(c  a)]=0⇒b2−2bc+c2[4(aca2bc+ab)]=0⇒b2−2bc+c24ac+4a2+4bc−4ab=0⇒b2+2bc+c2+4a24ac4ab=0⇒(b+c−2a)2=0⇒b + c = 2a$\begin{array}{l}\left(ab\right)x^2+\left(bc\right)x+\left(ca\right)=0\\ Given:2a=b+c\\ B^2-4AC=0\\ {\left(bc\right)}^2\left[4\left(ab\right)\left(ca\right)\right]=0\\ \Rightarrow b^2-2bc+c^2\left[4\left(ac{a}^{2}bc+ab\right)\right]=0\\ \Rightarrow b^2-2bc+c^{2}4ac+4a^2+4bc-\text{4}ab=0\\ \Rightarrow b^2+2bc+c^2+4a^{2}4ac4ab=0\\ \Rightarrow {\left(\text{b}+c-2a\right)}^2=0\\ \Rightarrow b+c=2a\end{array}$

Hence proved.

13. X and Y are centers of circles of radius 9cm and 2cm and XY = 17cm. Z is the center of a circle of radius 4 cm, which touches the above circles externally. Given that ∠XYZ=90∘$\mathrm{\angle }XYZ=90{}^{\circ }$, write an equation in r and solve it for r. 

Ans: 

Let r be the radius of the third circle 

XY = 17cm 

⇒ XZ = 9 + r 

    YZ = 2 + r

(r + 9)2+(r + 2)2=(17)2⇒r2+18r+81+r2+4r+4=289⇒r2+22r−204 =0⇒r2+11r−102 =0⇒(r+17)(r−6)=0⇒r=−17 (not possible) or r = 6 cm∴radius = 6cm.$\begin{array}{l}{\left(r+9\right)}^2+{\left(r+2\right)}^2={\left(17\right)}^2\\ \Rightarrow r^2+18r+81+r^2+4r+4=289\\ \Rightarrow r^2+22r-204=0\\ \Rightarrow r^2+11r-102=0\\ \Rightarrow \left(r+17\right)\left(r-6\right)=0\\ \Rightarrow r=-17\left(notpossible\right)orr=6cm\\ \therefore radius=6cm.\end{array}$

Level - 01 (01 Marks) 

1. Check whether the following are quadratic equation or not 

i. (x − 3) (2x + 1) = x(x + 5)$\left(x-3\right)\left(2x+1\right)=x\left(x+5\right)$

Ans:

Yes, this is a quadratic equation as the highest power of x is 2.

ii.  (x + 2)2= 2x(x2− 1)${\left(\mathbf{x}+\mathbf{2}\right)}^{\mathbf{2}}=\mathbf{2}\mathbf{x}({\mathbf{x}}^{\mathbf{2}}-\mathbf{1})$

Ans: No, this is not a quadratic equation as the highest power of x is 1.

2. Solve by factorization method x2−7x+12=0${\mathbf{x}}^{\mathbf{2}}-\mathbf{7}\mathbf{x}+\mathbf{12}=\mathbf{0}$

Ans: x = 3; x = 4$x=3;x=4$

3. Find the discriminant x2−3x−10= 0${\mathbf{x}}^{\mathbf{2}}-\mathbf{3}\mathbf{x}-\mathbf{10}=\mathbf{0}$

Ans: D = 49$D=49$ (D = discriminant)

4. Find the nature of root 2x2+ 3x − 4 = 0$\mathbf{2}{\mathbf{x}}^{\mathbf{2}}+\mathbf{3}\mathbf{x}-\mathbf{4}=\mathbf{0}$

Ans: root are real and unequal. 

5. Find the value k so that quadratic equation 3x2−kx+38=0$\mathbf{3}{\mathbf{x}}^{\mathbf{2}}-\mathbf{k}\mathbf{x}+\mathbf{38}=\mathbf{0}$ has equal root

Ans: 5 ± 18$5\pm 18$

6. Determine whether given value of x is a solution or not 

x2−3x−1=0:x = 1${\mathbf{x}}^{\mathbf{2}}-\mathbf{3}\mathbf{x}-\mathbf{1}=\mathbf{0}:\mathbf{x}=\mathbf{1}$

Ans: not a solution 

Level 2 (02 Marks)

7. Solve by quadratic equation 16x2− 24x − 1 = 0$\mathbf{16}{\mathbf{x}}^{\mathbf{2}}-\mathbf{24}\mathbf{x}-\mathbf{1}=\mathbf{0}$ by using quadratic formula. 

Ans: 1$1$

8. Determine the value of for which the quadratic equation 2x2+3x+k= 0$\mathbf{2}{\mathbf{x}}^{\mathbf{2}}+\mathbf{3}\mathbf{x}+\mathbf{k}=\mathbf{0}$ have both roots real.

Ans: 

k≤98x,x=3+10−−√4,3−10−−√4$k\le {\displaystyle \frac{9}{8}}x,x={\displaystyle \frac{3+\sqrt{10}}{4}},{\displaystyle \frac{3-\sqrt{10}}{4}}$

9. Find the roots of equation 2x2+x−6=0$\mathbf{2}{\mathbf{x}}^{\mathbf{2}}+\mathbf{x}-\mathbf{6}=\mathbf{0}$

Ans: x = 2,x=32$x=2,x={\displaystyle \frac{3}{2}}$

10. Find the roots of equation x−1x=3;x≠0$x-{\displaystyle \frac{1}{x}}=3;x\ne 0$

Ans:x=32$x={\displaystyle \frac{3}{2}}$ 

Level 3 (03 Marks)

1. The sum of the squares of two consecutive positive integers is 265. Find the integers. 

Ans: number are 11, 12 

2. Divide 39 into two parts such that their product is 324. 

Ans: 27, 12

3. The sum of the number and its reciprocals is. Find the number. 

Ans: 414$4{\displaystyle \frac{1}{4}}$ 

4. The length of a rectangle is 5cm more than its breadth if its area is 150 Sq. cm. 

Ans: 10cm, 15cm 

5. The altitude of a right triangle is 7cm less than its base. If the hypotenuse is 13cm. Find the other two sides. 

Ans: 12cm and 5cm 

1 Marks Questions 

1. Which of the following is a quadratic equation? 

a)x3−2x−5–√−x=0$a)x^3-2x-\sqrt{5}-x=0$ 

b)3x2−5x+9=x2−7x+3$b)3x^2-5x+9=x^2-7x+3$ 

c)(x+1x)2=3(x+1x)+4$c){\left(x+{\displaystyle \frac{1}{x}}\right)}^2=3\left(x+{\displaystyle \frac{1}{x}}\right)+4$ 

d)x3+x+3=0$d)x^3+x+3=0$ 

Ans: b)3x2−5x+9=x2−7x+3$b)3x^2-5x+9=x^2-7x+3$

2. Factor of a2x2−3abx+2b2=0$a^2{x}^2-3abx+2b^2=0$ is 

a)2ba,ba$a){\displaystyle \frac{2b}{a}},{\displaystyle \frac{b}{a}}$ 

b)3ba,ab$b){\displaystyle \frac{3b}{a}},{\displaystyle \frac{a}{b}}$ 

c)ba,ab$c){\displaystyle \frac{b}{a}},{\displaystyle \frac{a}{b}}$ 

d)ab,ab$d){\displaystyle \frac{a}{b}},{\displaystyle \frac{a}{b}}$ 

Ans: a)2ba,ba$a){\displaystyle \frac{2b}{a}},{\displaystyle \frac{b}{a}}$ 

3. Which of the following have real roots? 

a) 2x2+x−1=0$\text{a) 2}{\text{x}}^2+x-1=0$ 

b) x2+x+1=0$b){\text{x}}^2+x+1=0$ 

c) x2−6x+6=0$c){\text{x}}^2-6x+6=0$ 

d) 2x2+15x+30=0$d)\text{ 2}{\text{x}}^2+15x+30=0$ 

Ans: c) x2−6x+6=0$c){\text{x}}^2-6x+6=0$

4. Solve for x:

x=12−12−12−x$x={\displaystyle \frac{1}{2-{\displaystyle \frac{1}{2-{\displaystyle \frac{1}{2-x}}}}}}$

 a)x=2$a)x=2$ 

 b)x=−1$b)x=-1$ 

 c)x=1$c)x=1$ 

 d)x=3$d)x=3$ 

Ans: b)x=−1$b)x=-1$ 

5. Solve by factorization 3–√x2+10x+73–√=0$\sqrt{3}{x}^2+10x+7\sqrt{3}=0$

 a)x=−3–√,−73–√$a)x=-\sqrt{3},-{\displaystyle \frac{7}{\sqrt{3}}}$ 

 b)x=−3–√,73–√$b)x=-\sqrt{3},{\displaystyle \frac{7}{\sqrt{3}}}$ 

 c)x=2,12$c)x=2,{\displaystyle \frac{1}{2}}$ 

 d)±3$d)\pm 3$ 

Ans: a)x=−3–√,−73–√$a)x=-\sqrt{3},-{\displaystyle \frac{7}{\sqrt{3}}}$ 

6. The quadratic equation whose roots are 3 and -3 is 

 a)x2−9=0$a)x^2-9=0$ 

 b)x2−3x−3=0$b)x^2-3x-3=0$ 

 c)x2−2x+2=0$c)x^2-2x+2=0$ 

 d)x2+9=0$d)x^2+9=0$ 

Ans: a)x2−9=0$a)x^2-9=0$

7. Discriminant of −x2+12x+12=0$-x^2+{\displaystyle \frac{1}{2}}x+{\displaystyle \frac{1}{2}}=0$ is

a) −12,1$-{\displaystyle \frac{1}{2}},1$ 

b) 12,1${\displaystyle \frac{1}{2}},1$ 

c) −12,1${\displaystyle \frac{-1}{2}},1$ 

d) 12,−12${\displaystyle \frac{1}{2}},{\displaystyle \frac{-1}{2}}$ 

Ans: 

(a) −12,1$-{\displaystyle \frac{1}{2}},1$

8. For equal root,kx(x−2)+6=0$kx\left(x-2\right)+6=0$ , value of k is

a). k=6$k=6$ 

b). k=3$k=3$ 

c). k=2$k=2$ 

d. k=8$k=8$ 

Ans:

(a) k=6$k=6$

9. Quadratic equation whose roots are 2+s√,2−s√$2+\sqrt{s},2-\sqrt{s}$ is

a). x2−4x−1=0$x^2-4x-1=0$ 

b). x2+4x+1=0$x^2+4x+1=0$ 

c). x2+(x+5–√)x−(25–√)=0$x^2+\left(x+\sqrt{5}\right)x-\left(2\sqrt{5}\right)=0$ 

d). x2−4x+2=0$x^2-4x+2=0$ 

Ans: 

(a) x2−4x−1=0$x^2-4x-1=0$

10. If α$\alpha$ and β$\beta$ are roots of the equation 3x2+5x−7=0$3x^2+5x-7=0$ then αβ$\alpha \beta$ equal to

a). 73${\displaystyle \frac{7}{3}}$ 

b). −73${\displaystyle \frac{-7}{3}}$ 

c). −53${\displaystyle \frac{-5}{3}}$ 

d). 21$21$ 

Ans: 

(b) −73${\displaystyle \frac{-7}{3}}$

2 Marks Questions

1. Solve the following problems given-

i. x 2−45x+324=0$x{}^2-45x+324=0$ 

Ans:

x 2−45x+324=0$x{}^2-45x+324=0$

                    ⇒x2−36x−9x+324=0$\Rightarrow x^2-36x-9x+324=0$ 

                    ⇒x(x−36)−9(x−36)=0$\Rightarrow x\left(x-36\right)-9\left(x-36\right)=0$ 

    ⇒(x−9)(x−36)=0$\Rightarrow \left(x-9\right)\left(x-36\right)=0$ 

    ∴x=36,9$\therefore x=36,9$ 

ii. x 2−55x+750=0$x{}^2-55x+750=0$ 

Ans: 

                    ⇒x 2−25x−30x+750=0$\Rightarrow x{}^2-25x-30x+750=0$ 

                    ⇒ x(x−25)  30(x−25)=0 $\Rightarrow x\left(x-25\right)30\left(x-25\right)=0$ 

                    ⇒ (x−30)(x−25)=0$\Rightarrow \left(x-30\right)\left(x-25\right)=0$ 

                    ∴x=30,25$\therefore x=30,25$ 

2. Find two numbers whose sum is  27$27$ and the product is 182$182$ 

Ans: 

Let first number be x and let second number be (27−x)$\left(27-x\right)$ 

According to given condition, the product of two numbers is 182$182$.

Therefore,

      x(27−x)=182$x\left(27-x\right)=182$

      ⇒ 27x−x 2=182$\Rightarrow 27x-x{}^2=182$

      ⇒ x 2−27x+182=0$\Rightarrow x{}^2-27x+182=0$ 

              ⇒ x 2−27x+182=0$\Rightarrow x{}^2-27x+182=0$ 

              ⇒ x(x−14)  13(x−14)=0$\Rightarrow x\left(x-14\right)13\left(x-14\right)=0$ 

      ⇒ (x−14)(x−13)=0$\Rightarrow \left(x-14\right)\left(x-13\right)=0$ 

              ∴(x−14)(x−13)=0$\therefore \left(x-14\right)\left(x-13\right)=0$ 

       Therefore, the first number is equal to 14 or 13$14\text{ or 13}$ 

       And, second number is = 27  x=27  −14=13$=27x=27-14=13$ or Second number = 27  −13=14$=27-13=14$ 

       Therefore, two numbers are 13 and 14$13\text{ and }14$ 

3. Find two consecutive positive integers, the sum of whose squares is 365$365$.

Ans: 

Let first number be x and let second number be (x+1)$\left(x+1\right)$ 

According to given condition

x2+(x+1)2=365 {(a+b)2=a2+b 2+2ab}$x^2+{\left(x+1\right)}^2=365\left\{{\left(a+b\right)}^2=a^2+b{}^2+2ab\right\}$ 

⇒ x2+x2+1+2x=365$\Rightarrow x^2+x^2+1+2x=365$ 

⇒ 2x2+2x −364=0$\Rightarrow 2x^2+2x-364=0$ 

Dividing equation by 2$2$ 

⇒ x2+x −182=0$\Rightarrow x^2+x-182=0$ 

⇒ x2+14x−13x −182=0$\Rightarrow x^2+14x-13x-182=0$ 

     ⇒ x(x+14)  −13(x+14)=0$\Rightarrow x\left(x+14\right)-13\left(x+14\right)=0$ 

   ⇒ (x+14)(x−13)=0$\Rightarrow \left(x+14\right)\left(x-13\right)=0$ 

    ∴x=13,−14$\therefore x=13,-14$ 

Therefore, first number =13$=13$ {We discard −14$-14$ because it is negative number)
Second number = x+1=13+1=14$=x+1=13+1=14$ 

Therefore, two consecutive positive integers are 13and 14$13\text{and 14}$ whose sum of squares is equal to 365$365$.

4. The altitude of a right triangle is 7$7$cm less than its base. If, hypotenuse is 13$13$cm. Find the other two sides.

Ans: 

Let base of triangle be x cm and let altitude of triangle be (x−7) cm

It is given that hypotenuse of triangle is 13 cm

According to Pythagoras Theorem,

132 =x2+(x−7)2(a+b)2=a2+b2+2ab$132=x^2+{\left(x-7\right)}^2{\left(a+b\right)}^2=a^2+b^2+2ab$ 

⇒ 169=x2+x2+49−14x$\Rightarrow 169=x^2+x^2+49-14x$ 

⇒ 169=2x2−14x+49 $\Rightarrow 169=2x^2-14x+49$ 

⇒ 2x2−14x 120=0$\Rightarrow 2x^2-14x120=0$ 

Dividing equation by 2$2$ 

⇒ x2−7x 60=0$\Rightarrow x^2-7x60=0$ 

⇒ x2−12x+5x 60=0$\Rightarrow x^2-12x+5x60=0$ 

⇒ x(x−12)+5(x−12)=0$\Rightarrow x\left(x-12\right)+5\left(x-12\right)=0$ 

⇒ (x−12)(x+5)$\Rightarrow \left(x-12\right)\left(x+5\right)$ 

∴x=−5,12 $\therefore x=-5,12$ 

We discard x=−5$x=-5$ because the length of the side of the triangle cannot be negative.

Therefore, the base of triangle =12$=12$cm

Altitude of triangle =(x−7)=12−7=5$=\left(x-7\right)=12-7=5$ cm

5. A cottage industry produces a certain number of pottery articles in a day. It was observed on a particular day that the cost of production of each article (in rupees) was 3$3$ more than twice the number of articles produced on that day. If, the total cost of production on that day was Rs.90$90$ , find the number of articles produced and the cost of each article.

Ans:  

Let cost of production of each article be Rs x$x$ 

We are given total cost of production on that particular day = Rs 90$=Rs90$ 

Therefore, total number of articles produced that day = 90x$={\displaystyle \frac{90}{x}}$ 

According to the given conditions,

x=2(90x)+3$x=2\left({\displaystyle \frac{90}{x}}\right)+3$ 

⇒x=180x+3$\Rightarrow x={\displaystyle \frac{180}{x}}+3$ 

⇒x=180+3xx$\Rightarrow x={\displaystyle \frac{180+3x}{x}}$ 

⇒ x2=180+3x$\Rightarrow x^2=180+3x$ 

⇒ x2−3x 180=0$\Rightarrow x^2-3x180=0$ 

⇒ x2−15x+12x 180=0$\Rightarrow x^2-15x+12x180=0$ 

⇒ x(x−15)+12(x−15)=0$\Rightarrow x\left(x-15\right)+12\left(x-15\right)=0$ 

    ⇒ (x−15)(x+12)=0$\Rightarrow \left(x-15\right)\left(x+12\right)=0$     

     ∴x=15,−12$\therefore x=15,-12$      

Cost cannot be in negative; therefore, we discard x=−12$x=-12$ 

Therefore, x=Rs15$,x=Rs15$which is the cost of production of each article.

Number of articles produced on that particular day      =9015= 6$={\displaystyle \frac{90}{15}}=6$ 

6. In a class test, the sum of Shefali's marks in Mathematics and English is 30$30$. Had she got 2$2$ marks more in Mathematics and 3$3$ marks less in English, the product of their marks would have been210$210$. Find her marks in the two subjects.

Ans: 

Let Shefali's marks in Mathematics = x$=x$ 

Let Shefali's marks in English = 30−x$=30-x$ 

If, she had got 2 marks more in Mathematics, her marks would be = x+2$=x+2$ 

If, she had got 3 marks less in English, her marks in English would be = 30  x−3 = 27−x$=30x-3=27-x$ 

According to given condition:

(x+2)(27−x)=210$\left(x+2\right)\left(27-x\right)=210$ 

⇒ 27x−x2+54−2x=210$\Rightarrow 27x-x^2+54-2x=210$ 

⇒ x2−25x+156=0$\Rightarrow x^2-25x+156=0$ 

Comparing quadratic equation x2−25x+156=0$x^2-25x+156=0$with general formax2+bx+c=0$a{x}^2+bx+c=0$,

We get a=1,b=−25 and c=156$a=1,b=-25andc=156$ 

Applying Quadratic Formulax=−b±b2−4ac−−−−−−−√2a$x={\displaystyle \frac{-b\pm \sqrt{b^2-4ac}}{2a}}$ 

x=25±(25)2−4(1)(156)−−−−−−−−−−−−−−√2×1$x={\displaystyle \frac{25\pm \sqrt{{\left(25\right)}^2-4\left(1\right)\left(156\right)}}{2\times 1}}$ 

⇒x=25±625−624−−−−−−−−√2$\Rightarrow x={\displaystyle \frac{25\pm \sqrt{625-624}}{2}}$ 

x=25±1–√2$x={\displaystyle \frac{25\pm \sqrt{1}}{2}}$ 

⇒x=25+12,25−12$\Rightarrow x={\displaystyle \frac{25+1}{2}},{\displaystyle \frac{25-1}{2}}$ 

∴x=13,12$\therefore x=13,12$ 

Therefore, Shefali's marks in Mathematics = 13 or 12$=13or12$ 

Shefali's marks in English = 30  x=30  13=17$=30x=3013=17$ 

Or Shefali's marks in English = 30  x=30  12=18$=30x=3012=18$ 

Therefore, her marks in Mathematics and English are (13,17) or (12,18).$\left(13,17\right)or\left(12,18\right).$ 

7. The diagonal of a rectangular field is 60$60$ meters more than the shorter side. If, the longer side is 30$30$ meters more than the shorter side, find the sides of the field.

Ans:

Let shorter side of rectangle = x$=x$meters

Let diagonal of rectangle = (x+60)$=\left(x+60\right)$meters 

Let longer side of rectangle = (x+30)$=\left(x+30\right)$meters 

According to Pythagoras theorem,

(x+60)2=(x+30)2+x2${\left(x+60\right)}^2={\left(x+30\right)}^2+x^2$ 

⇒ x2+3600+120x=x2+900+60x+x2$\Rightarrow x^2+3600+120x=x^2+900+60x+x^2$ 

 ⇒ x2−60x 2700=0$\Rightarrow x^2-60x2700=0$ 

x2−60x 2700=0$x^2-60x2700=0$ 

Comparing equation x2−60x 2700=0$x^2-60x2700=0$with standard formax2+bx+c=0$a{x}^2+bx+c=0$, 

We get a=1,b=−60 and c=−2700$a=1,b=-60andc=-2700$ 

Applying quadratic formula x=−b±b2−4ac−−−−−−−√2a$x={\displaystyle \frac{-b\pm \sqrt{b^2-4ac}}{2a}}$ 

x=60±(60)2−4(1)(−2700)−−−−−−−−−−−−−−−−−√2×1$x={\displaystyle \frac{60\pm \sqrt{{\left(60\right)}^2-4\left(1\right)\left(-2700\right)}}{2\times 1}}$ 

⇒x=60±3600+10800−−−−−−−−−−−√2$\Rightarrow x={\displaystyle \frac{60\pm \sqrt{3600+10800}}{2}}$ 

⇒x=60±14400−−−−−√2=60±1202$\Rightarrow x={\displaystyle \frac{60\pm \sqrt{14400}}{2}}={\displaystyle \frac{60\pm 120}{2}}$ 

⇒x=60+1202,60−1202$\Rightarrow x={\displaystyle \frac{60+120}{2}},{\displaystyle \frac{60-120}{2}}$ 

∴x=90,−30$\therefore x=90,-30$ 

We ignore −30$-30$ . Since length cannot be negative. 

Therefore, x=90$x=90$ which means length of shorter side =90$=90$ meters 

And length of longer side = x+30 = 90+30=120$=x+30=90+30=120$meters 

Therefore, length of sides is 90 and 120$90\text{ and }120$ in meters.

8. The difference of squares of two numbers is 180$180$. The square of the smaller number is 8$8$ times the larger number. Find the two numbers.

Ans: 

Let smaller number = x$=x$and let larger number = y$=y$ 

According to condition: 

y2−x2=180 … (1)$y^2-x^2=180\dots \left(1\right)$ 

Also, we are given that square of smaller number is 8$8$ times the larger number. 

⇒ x2=8y … (2)$\Rightarrow x^2=8y\dots \left(2\right)$ 

Putting equation (2) in (1), we get

 y2−8y=180$y^2-8y=180$ 

⇒ y2−8y 180=0$\Rightarrow y^2-8y180=0$ Comparing equation y2−8y 180=0$y^2-8y180=0$with general form ay2+by+c=0$a{y}^2+by+c=0$, 

We get a=1,b=−8 and c=−180$a=1,b=-8\text{ and }c=-180$ 

Using quadratic formula y=−b±b2−4ac−−−−−−−√2a$y={\displaystyle \frac{-b\pm \sqrt{b^2-4ac}}{2a}}$ 

y=8±(−8)2−4(1)(−180)−−−−−−−−−−−−−−−−√2×1$y={\displaystyle \frac{8\pm \sqrt{{\left(-8\right)}^2-4\left(1\right)\left(-180\right)}}{2\times 1}}$ 

⇒y=8±64+720−−−−−−−√2$\Rightarrow y={\displaystyle \frac{8\pm \sqrt{64+720}}{2}}$ 

⇒y=8±784−−−√2=8±282$\Rightarrow y={\displaystyle \frac{8\pm \sqrt{784}}{2}}={\displaystyle \frac{8\pm 28}{2}}$ 

⇒y=8+282,8−282$\Rightarrow y={\displaystyle \frac{8+28}{2}},{\displaystyle \frac{8-28}{2}}$ 

∴y=18,−10$\therefore y=18,-10$ 

Using equation (2) to find smaller number:

x 2 =8y$x{}^2=8y$ 

⇒ x2=8y=8×18=144$\Rightarrow x^2=8y=8\times 18=144$ 

     ⇒ x=±12$\Rightarrow x=\pm 12$ 

     And,x2=8y=8×−10=−80$x^2=8y=8\times -10=-80$  {No real solution for x$x$} 

  Therefore, two numbers are (12,18) or (−12,18)$\left(12,18\right)or\left(-12,18\right)$ 

9. A train travels 360 km at a uniform speed. If, the speed had been 5 km/hr. more, it would have taken 1 hour less for the same journey. Find the speed of the train.

Ans: 

Let the speed of the train = x km/hr 

If, speed had been 5km/hr more, train would have taken 1 hour less. 

So, according to this condition

360x=360x+5+1${\displaystyle \frac{360}{x}}={\displaystyle \frac{360}{x+5}}+1$ 

⇒360(1x−1x+5)=1$\Rightarrow 360\left({\displaystyle \frac{1}{x}}-{\displaystyle \frac{1}{x+5}}\right)=1$ 

⇒360(x+5−xx(x+5))=1$\Rightarrow 360\left({\displaystyle \frac{x+5-x}{x\left(x+5\right)}}\right)=1$ 

⇒ 360×5=x2+5x$\Rightarrow 360\times 5=x^2+5x$ 

⇒ x2+5x 1800=0$\Rightarrow x^2+5x1800=0$ 

Comparing equation x2+5x 1800=0$x^2+5x1800=0$ with general equationax2+bx+c=0$a{x}^2+bx+c=0$ , 

We get a=1,b=5 and c=−1800$a=1,b=5\text{ and }c=-1800$ 

Applying quadratic formula x=−b±b2−4ac−−−−−−−√2a$x={\displaystyle \frac{-b\pm \sqrt{b^2-4ac}}{2a}}$ 

x=−5±(5)2−4(1)(−1800)−−−−−−−−−−−−−−−−√2×1$x={\displaystyle \frac{-5\pm \sqrt{{\left(5\right)}^2-4\left(1\right)\left(-1800\right)}}{2\times 1}}$ 

⇒x=−5±25+7200−−−−−−−−√2$\Rightarrow x={\displaystyle \frac{-5\pm \sqrt{25+7200}}{2}}$ 

⇒x=−5±7225−−−−√2=−5±852$\Rightarrow x={\displaystyle \frac{-5\pm \sqrt{7225}}{2}}={\displaystyle \frac{-5\pm 85}{2}}$ 

⇒x=40,−45$\Rightarrow x=40,-45$ 

Since the speed of train cannot be in negative. Therefore, we discard x=−45$x=-45$ 

Therefore, speed of train = 40$=40$km/hr

10. Find the value of k for each of the following quadratic equations, so that they have two equal roots.

i. 2x2+kx+3=0$2x^2+kx+3=0$ 

Ans:

2x2+kx+3=0$2x^2+kx+3=0$

We know that a quadratic equation has two equal roots only when the value of the discriminant is equal to zero.                     

Comparing equation 2x2+kx+3=0$2x^2+kx+3=0$ with general quadratic equationax2+bx+c=0$a{x}^2+bx+c=0$,

  we get a=2,b=k and c=3$a=2,b=kandc=3$ 

  Discriminant = b2−4ac=k2 4(2)(3)=k2−24$=b^2-4ac=k^{2}4\left(2\right)\left(3\right)=k^2-24$

  Putting discriminant equal to zero

 k2 24=0$k^{2}24=0$ 

⇒k2=24$\Rightarrow k^2=24$ 

⇒k=±24−−√=±26–√$\Rightarrow k=\pm \sqrt{24}=\pm 2\sqrt{6}$ 

⇒k=26–√,−26–√$\Rightarrow k=2\sqrt{6},-2\sqrt{6}$ 

ii. kx(x−2)+6=0$kx\left(x-2\right)+6=0$ 

Ans: 

kx(x−2)+6=0$kx\left(x-2\right)+6=0$ 

⇒ kx2−2kx+6=0$\Rightarrow k{x}^2-2kx+6=0$ 

Comparing quadratic equation kx2−2kx+6=0$k{x}^2-2kx+6=0$ with general formax2+bx+c=0$a{x}^2+bx+c=0$, we get a=k,b= −2k andc=6$a=k,b=-2kandc=6$ 

Discriminant = b2−4ac=(−2k)2 4(k)(6)=4k2−24k$=b^2-4ac={\left(-2k\right)}^{2}4\left(k\right)\left(6\right)=4k^2-24k$ 

We know that two roots of quadratic equation are equal only if discriminant is equal to zero. 

Putting discriminant equal to zero

4k2−24k=0$4k^2-24k=0$  

⇒ 4k(k−6)=0$\Rightarrow 4k\left(k-6\right)=0$ 

⇒ k=0,6$\Rightarrow k=0,6$ 

The basic definition of quadratic equation says that quadratic equation is the equation of the formax2+bx+c=0$a{x}^2+bx+c=0$ , where a≠0.$a\ne 0.$ 

Therefore, in equationkx2−2kx+6=0$k{x}^2-2kx+6=0$, we cannot have k=0$k=0$. 

Therefore, we discard k=0$k=0$. 

Hence the answer is k=6$k=6$ 

11.  Is it possible to design a rectangular mango grove whose length is twice its breadth, and the area is 800 m2$800m^2$. If so, find its length and breadth.

Ans: 

Let breadth of rectangular mango grove = x$=x$meters 

Let length of rectangular mango grove = 2x$=2x$ meters 

Area of rectangle = length × breadth = x× 2x = 2x2m2$=x\times 2x=2x^2{m}^2$ 

According to given condition-

2x2=800$2x^2=800$ 

⇒ 2x2 800=0$\Rightarrow 2x^{2}800=0$ 

⇒ x2 400=0$\Rightarrow x^{2}400=0$ 

Comparing equation x2 400=0$x^{2}400=0$with general form of quadratic equationax2+bx+c=0$a{x}^2+bx+c=0$, we get a=1,b=0 and c=400$a=1,b=0\text{ and }c=400$

Discriminant = b2−4ac=(0)2 4(1)(−400)=1600$=b^2-4ac={\left(0\right)}^{2}4\left(1\right)\left(-400\right)=1600$ 

Discriminant is greater than 0 means that equation has two distinct real roots.

Therefore, it is possible to design a rectangular grove.

Applying quadratic formula, x=−b±b2−4ac−−−−−−−√2a$x={\displaystyle \frac{-b\pm \sqrt{b^2-4ac}}{2a}}$ to solve equation,

x=0±1600−−−−√2×1=±402=±20$x={\displaystyle \frac{0\pm \sqrt{1600}}{2\times 1}}={\displaystyle \frac{\pm 40}{2}}=\pm 20$ 

∴x=20,−20$\therefore x=20,-20$ We discard negative value of x$x$ because breadth of rectangle cannot be in negative. 

Therefore, x =$x=$ breadth of rectangle = 20$=20$ meters 

Length of rectangle = 2x=2×20=40$=2x=2\times 20=40$ meters

12. Is the following situation possible? If so, determine their present ages. The sum of the ages of two friends is  20$20$ years. Four years ago, the product of their ages in years was 48$48$.

Ans: 

Let age of first friend = x years and let age of second friend = (20−x)$=\left(20-x\right)$ years 

Four years ago, age of first friend = (x−4)$=\left(x-4\right)$ years 

Four years ago, age of second friend = (20−x)−4 = (16−x)$=\left(20-x\right)-4=\left(16-x\right)$ years 

According to given condition,

(x−4)(16−x)=48$\left(x-4\right)\left(16-x\right)=48$ 

 ⇒ 16x−x2 64+4x=48$\Rightarrow 16x-x^{2}64+4x=48$  

⇒ 20x−x2 112=0$\Rightarrow 20x-x^{2}112=0$ 

⇒ x2−20x+112=0$\Rightarrow x^2-20x+112=0$ 

Comparing equation, x2−20x+112=0$x^2-20x+112=0$ with general quadratic equationax2+bx+c=0$a{x}^2+bx+c=0$, we get a=1,b=−20 and c=112$a=1,b=-20andc=112$ 

Discriminant =b2−4ac=(−20)2 4(1)(112)=400  448=−48<0$=b^2-4ac={\left(-20\right)}^{2}4\left(1\right)\left(112\right)=400448=-48<0$ 

The discriminant is less than zero which means we have no real roots for this equation.

Therefore, the given situation is not possible.

13. Value ofx$x$  for x2−8x+15=0$x^2-8x+15=0$ is quadratic formula is

a). 3, 2$3,2$ 

b). 5, 2$5,2$ 

c). 5, 3$5,3$ 

d). 2, 3$2,3$ 

Ans: 

(c) 5, 3$5,3$

14. Discriminate of 3–√x2−22–√x−23–√=0$\sqrt{3}{x}^2-2\sqrt{2}x-2\sqrt{3}=0$ is

a). 30$30$ 

b). 31$31$ 

c). 32$32$ 

d). 35$35$ 

Ans: 

(c) 32$32$

15. Solve 12abx2−9a2x+8b2x−6ab=0$12ab{x}^2-9a^{2}x+8b^{2}x-6ab=0$ 

Ans: 

12abx2−9a2x+8b2x−6ab=0$12ab{x}^2-9a^{2}x+8b^{2}x-6ab=0$

⇒3ax(4bx−3x)+2b(4bx−3x)=0$\Rightarrow 3ax\left(4bx-3x\right)+2b\left(4bx-3x\right)=0$ 

⇒(4bx−3x)(3ax+2b)=0$\Rightarrow \left(4bx-3x\right)\left(3ax+2b\right)=0$ 

⇒4bx−3a=0or3ax+2b=0$\Rightarrow 4bx-3a=0or3ax+2b=0$ 

∴x=3a4borx=−2b3a$\therefore x={\displaystyle \frac{3a}{4b}}orx=-{\displaystyle \frac{2b}{3a}}$ 

16. Solve for x$x$ by quadratic formulap2x2+(p2−q2)x−q2=0$p^2{x}^2+\left(p^2-q^2\right)x-q^2=0$ 

Ans: 

p2x2+(p2−q2)x−q2=0$p^2{x}^2+\left(p^2-q^2\right)x-q^2=0$

a=p2,b=p2−q2,c=−q2$a=p^2,b=p^2-q^2,c=-q^2$ 

D=b2−4ac$D=b^2-4ac$ 

=(p2−q2)−4×p2(−q2)$=\left(p^2-q^2\right)-4\times p^2\left(-q^2\right)$ 

=p4+q2−2p2q2+4p2q2$=p^4+q^2-2p^2{q}^2+4p^2{q}^2$ 

=(p2+q2)2$={\left(p^2+q^2\right)}^2$ 

x=−b±D−−√29$x={\displaystyle \frac{-b\pm \sqrt{D}}{29}}$ 

=−(p2q2)±(p2+q2)2−−−−−−−−√2×p2$={\displaystyle \frac{-\left(p^2{q}^2\right)\pm \sqrt{{\left(p^2+q^2\right)}^2}}{2\times p^2}}$ 

=−p2+q2+p2+q22p2$={\displaystyle \frac{-p^2+q^2+p^2+q^2}{2p^2}}$ 

orx=−p2+q2−p22p2$orx={\displaystyle \frac{-p^2+q^2-p^2}{2p^2}}$ x=2q22p2orx=−2q22p2$x={\displaystyle \frac{2q^2}{2p^2}}orx={\displaystyle \frac{-2q^2}{2p^2}}$ 

x=q2p2orx=−1$x={\displaystyle \frac{q^2}{p^2}}orx=-1$ 

17. Find the value of k for which the quadratic equationkx2+2x+1=0$k{x}^2+2x+1=0$  has real and 

distinct root

Ans: 

kx2+2x+1=0$k{x}^2+2x+1=0$

a=k,b=2,c=1$a=k,b=2,c=1$ 

b=b2−4ac$b=b^2-4ac$ 

=(2)2−4×k×1=4−4k$={\left(2\right)}^2-4\times k\times 1=4-4k$ 

For real and distinct roots,

D>0$D>0$ 

4−4k>0$4-4k>0$ 

⇒−4k>−4$\Rightarrow -4k>-4$ 

∴k<1$\therefore k<1$ 

18. If one root of the equations 2x2+ax+3=0$2x^2+ax+3=0$ is 1$1$ , find the value of a.

a). =−4$=-4$ 

b). =−5$=-5$ 

c). =−3$=-3$ 

d). =−1$=-1$ 

Ans: 

(b) =−5$=-5$

19. Find k for which the quadratic equation4x2−3kx+1=0$4x^2-3kx+1=0$  has equal root.

1. =±34$=\pm {\displaystyle \frac{3}{4}}$ 

2. =34$={\displaystyle \frac{3}{4}}$ 

3. =±43$=\pm {\displaystyle \frac{4}{3}}$ 

4. =23$={\displaystyle \frac{2}{3}}$ 

Ans: 

(c) =±43$=\pm {\displaystyle \frac{4}{3}}$

20. Determine the nature of the roots of the quadratic equation

9a2b2x2−24abcdx+16c2d2=0$9a^2{b}^2{x}^2-24abcdx+16c^2{d}^2=0$ 

Ans:

D=b2−4ac$D=b^2-4ac$ 

=(−24abcd)2−4×9a2b2×16c2d2$={\left(-24abcd\right)}^2-4\times 9a^2{b}^2\times 16c^2{d}^2$ 

=576a2b2c2−376a2b2c2d2=0$=576a^2{b}^2{c}^2-376a^2{b}^2{c}^2{d}^2=0$ 

21. Find the discriminant of the equation (x−1)(2x−1)=0$\left(x-1\right)\left(2x-1\right)=0$ 

Ans: 

(x−1)(2x−1)=0$\left(x-1\right)\left(2x-1\right)=0$

⇒2x2−x−2x+1=0$\Rightarrow 2x^2-x-2x+1=0$

 ⇒2x2−3x+1=0$\Rightarrow 2x^2-3x+1=0$ 

Here,a=2,b=−3,c=1$Here,a=2,b=-3,c=1$ 

D=b2−4ac$D=b^2-4ac$ 

=(−3)2−4×2×1$={\left(-3\right)}^2-4\times 2\times 1$ 

=9−8=1$=9-8=1$ 

22. Find the value of k so that (x−1)$\left(x-1\right)$ is a factor of k2x2−2kx−3$k^2{x}^2-2kx-3$.

Ans: 

Let P(x)=k2x2−2kx−3$P\left(x\right)=k^2{x}^2-2kx-3$

P(1)=k2(1)2−2k(1)−3$P\left(1\right)=k^2{\left(1\right)}^2-2k\left(1\right)-3$ 

⇒0=k2−2k−3$\Rightarrow 0=k^2-2k-3$ 

⇒k2−3k+k−3$\Rightarrow k^2-3k+k-3$ 

⇒k(k−3)+1(k−3)=0$\Rightarrow k\left(k-3\right)+1\left(k-3\right)=0$ 

⇒(k−3)(k+1)=0$\Rightarrow \left(k-3\right)\left(k+1\right)=0$ 

∴k=3ork=−1$\therefore k=3ork=-1$ 

23. The product of two consecutive positive integers is 306$306$. Represent these in quadratic 

equation.

a). x2+x−306=0$x^2+x-306=0$

b). x2−x−306=0$x^2-x-306=0$

c). x2+2x−106=0$x^2+2x-106=0$ 

d). x2−x−106=0$x^2-x-106=0$ 

Ans: 

(a) x2+x−306=0$x^2+x-306=0$

24. Which is a quadratic equation?

a). x2+x+2=0$x^2+x+2=0$ 

b). x3+x2+2=0$x^3+x^2+2=0$ 

c). x4+x2+2=0$x^4+x^2+2=0$ 

d). x+2=0$x+2=0$ 

Ans: 

(a) x2+x+2=0$x^2+x+2=0$

25. The sum of two numbers is 16$16$. The sum of their reciprocals is 13${\displaystyle \frac{1}{3}}$. Find the numbers.  

Ans: 

Let no. be x$x$ 

According to question,

1x+116−x=13${\displaystyle \frac{1}{x}}+{\displaystyle \frac{1}{16-x}}={\displaystyle \frac{1}{3}}$ 

⇒1616x−x2=13$\Rightarrow {\displaystyle \frac{16}{16x-x^2}}={\displaystyle \frac{1}{3}}$ 

⇒x2−16x+48=0$\Rightarrow x^2-16x+48=0$ 

⇒x2−12x−4x+48=0$\Rightarrow x^2-12x-4x+48=0$ 

∴x=12orx=4$\therefore x=12orx=4$ 

26. Solve for x:217−x−−−−−−√=x−7$x:\sqrt{217-x}=x-7$ 

Ans: 

217−x−−−−−−√=(x−7)$\sqrt{217-x}=\left(x-7\right)$

⇒217−x=x2+49−14x$\Rightarrow 217-x=x^2+49-14x$ 

⇒x2−14x+x+49−217=0$\Rightarrow x^2-14x+x+49-217=0$ 

⇒x2−13x−168=0$\Rightarrow x^2-13x-168=0$ 

⇒x2−21x+8x−168=0$\Rightarrow x^2-21x+8x-168=0$ 

∴x=21orx=−8$\therefore x=21orx=-8$ 

27. Solve for x by factorization: x+1x=11111$x+{\displaystyle \frac{1}{x}}=11{\displaystyle \frac{1}{11}}$ 

Ans:

x2+1x=12211${\displaystyle \frac{x^2+1}{x}}={\displaystyle \frac{122}{11}}$ 

⇒11x2−12x+11=0$\Rightarrow 11x^2-12x+11=0$ 

⇒11x2−121x−1x+11=0$\Rightarrow 11x^2-121x-1x+11=0$ 

⇒11x(x−11)−1(x−11)=0$\Rightarrow 11x\left(x-11\right)-1\left(x-11\right)=0$ 

⇒(11x−1)(x−11)=0$\Rightarrow \left(11x-1\right)\left(x-11\right)=0$ 

∴x=11orx=111$\therefore x=11orx={\displaystyle \frac{1}{11}}$ 

28. Find the ratio of the sum and product of the roots of 7x2−12x+18=0$7x^2-12x+18=0$ 

Ans:  

7x2−12x+18=0$7x^2-12x+18=0$

α+β=−ba=127andαβ=ca=1817$\alpha +\beta ={\displaystyle \frac{-b}{a}}={\displaystyle \frac{12}{7}}and\alpha \beta ={\displaystyle \frac{c}{a}}={\displaystyle \frac{18}{17}}$

α+βαβ=1271817=127×1718=3421${\displaystyle \frac{\alpha +\beta }{\alpha \beta }}={\displaystyle \frac{{\displaystyle \frac{12}{7}}}{{\displaystyle \frac{18}{17}}}}={\displaystyle \frac{12}{7}}\times {\displaystyle \frac{17}{18}}={\displaystyle \frac{34}{21}}$ 

29. If α$\alpha$  andβ$\beta$  are the roots of the equation x2+kx+12=0$x^2+kx+12=0$, such that α−β=1$\alpha -\beta =1$ , then

Ans:

α+β=−k1,$\alpha +\beta ={\displaystyle \frac{-k}{1}},$ 

α−β=1$\alpha -\beta =1$ 

αβ=121$\alpha \beta ={\displaystyle \frac{12}{1}}$ 

(α+β)2=(α−β)2+4αβ${\left(\alpha +\beta \right)}^2={\left(\alpha -\beta \right)}^2+4\alpha \beta$ 

⇒(−k)2=(1)2+4×12$\Rightarrow {\left(-k\right)}^2={\left(1\right)}^2+4\times 12$ 

⇒k2=49$\Rightarrow k^2=49$ 

k=±7$k=\pm 7$ 

3 Marks Questions

1. Check whether the following are Quadratic Equations.

i). (x+1)2=2(x−3)${\left(x+1\right)}^2=2\left(x-3\right)$ 

Ans: 

(x+1)2=2(x−3){(a+b)2=a2+2ab+b2 }${\left(x+1\right)}^2=2\left(x-3\right)\left\{{\left(a+b\right)}^2=a^2+2ab+b^2\right\}$ 

⇒ x2+1+2x=2x 6$\Rightarrow x^2+1+2x=2x6$ 

⇒ x2+7=0$\Rightarrow x^2+7=0$ 

Here, degree of equation is 2$2$.

Therefore, it is a Quadratic Equation.

ii). x2−2x=(−2)(3−x)$x^2-2x=\left(-2\right)\left(3-x\right)$ 

Ans:

x2−2x=(−2)(3−x)$x^2-2x=\left(-2\right)\left(3-x\right)$

⇒ x2−2x=−6+2x$\Rightarrow x^2-2x=-6+2x$

⇒ x2−2x−2x+6=0$\Rightarrow x^2-2x-2x+6=0$ 

⇒ x2−4x+6=0$\Rightarrow x^2-4x+6=0$ 

Here, degree of equation is 2.$2.$ 

Therefore, it is a Quadratic Equation.

iii). (x−2)(x+1)=(x−1)(x+3)$\left(x-2\right)\left(x+1\right)=\left(x-1\right)\left(x+3\right)$ 

Ans:

(x−2)(x+1)=(x−1)(x+3)

⇒ x2+x−2x 2=x2+3x x 3=0$\Rightarrow x^2+x-2x2=x^2+3xx3=0$ 

⇒ x2+x−2x 2−x2−3x+x+3=0$\Rightarrow x^2+x-2x2-x^2-3x+x+3=0$ 

⇒ x−2x 2−3x+x+3=0$\Rightarrow x-2x2-3x+x+3=0$ 

⇒ −3x+1=0$\Rightarrow -3x+1=0$ 

Here, degree of equation is 1.$1.$ 

Therefore, it is not a Quadratic Equation.

iv).   (x−3)(2x+1)=x(x+5)$\left(x-3\right)\left(2x+1\right)=x\left(x+5\right)$

Ans:

(x−3)(2x+1)=x(x+5)$\left(x-3\right)\left(2x+1\right)=x\left(x+5\right)$ 

⇒ 2x2+x−6x 3=x2+5x$\Rightarrow 2x^2+x-6x3=x^2+5x$ 

⇒ 2x2+x−6x 3−x2−5x=0$\Rightarrow 2x^2+x-6x3-x^2-5x=0$

⇒ x2−10x 3=0$\Rightarrow x^2-10x3=0$

Here, degree of equation is 2.

Therefore, it is a quadratic equation.

v). (2x−1)(x−3)=(x+5)(x−1)$\left(2x-1\right)\left(x-3\right)=\left(x+5\right)\left(x-1\right)$ 

Ans:

(2x−1)(x−3)=(x+5)(x−1) $\left(2x-1\right)\left(x-3\right)=\left(x+5\right)\left(x-1\right)$

⇒ 2x2−6x x+3=x2 x+5x 5$\Rightarrow 2x^2-6xx+3=x^{2}x+5x5$ 

⇒ x2−11x+8=0$\Rightarrow x^2-11x+8=0$

Here, degree of Equation is2$2$.

Therefore, it is a Quadratic Equation.

vi). x2+3x+1=(x−2)2$x^2+3x+1={\left(x-2\right)}^2$ 

Ans:

x2+3x+1=(x−2)2 {(a−b)2=a2−2ab+b2}$x^2+3x+1={\left(x-2\right)}^2\left\{{\left(a-b\right)}^2=a^2-2ab+b^2\right\}$ 

⇒ x2 +3x+1=x2+4−4x$\Rightarrow x^2+3x+1=x^2+4-4x$ 

⇒ x2+3x+1−x2+4x 4=0$\Rightarrow x^2+3x+1-x^2+4x4=0$

⇒ 7x 3=0$\Rightarrow 7x3=0$

Here, degree of equation is 1$1$.

Therefore, it is not a Quadratic Equation.

vii). (x+2)3=2x(x2−1)${\left(x+2\right)}^3=2x\left(x^2-1\right)$ 

Ans:

(x+2)3=2x(x2−1) {(a+b)3=a3+b3+3ab(a+b)}${\left(x+2\right)}^3=2x\left(x^2-1\right)\left\{{\left(a+b\right)}^3=a^3+b^3+3ab\left(a+b\right)\right\}$

⇒ x3 +23 +3(x)(2)(x+2)=2x(x2−1)$\Rightarrow x^3+2^3+3\left(x\right)\left(2\right)\left(x+2\right)=2x\left(x^2-1\right)$

⇒ x3+8+6x(x+2)=2x3−2x$\Rightarrow x^3+8+6x\left(x+2\right)=2x^3-2x$

⇒ 2x3−2x−x3 8−6x2−12x=0$\Rightarrow 2x^3-2x-x^{3}8-6x^2-12x=0$

⇒ x3−6x2−14x 8=0$\Rightarrow x^3-6x^2-14x8=0$

Here, degree of Equation is 3.

Therefore, it is not a quadratic Equation.

viii). x3−4x2 x+1=(x−2)3 $x^3-4x^{2}x+1={\left(x-2\right)}^3$

Ans:

x3−4x2 x+1=(x−2)3 {(a−b)3 =a3−b3−3ab(a−b)}$x^3-4x^{2}x+1={\left(x-2\right)}^3\left\{{\left(a-b\right)}^3=a^3-b^3-3ab\left(a-b\right)\right\}$ 

   ⇒ x3−4x2 x+1=x3−23 3(x)(2)(x−2)$\Rightarrow x^3-4x^{2}x+1=x^3-2^{3}3\left(x\right)\left(2\right)\left(x-2\right)$ 

   ⇒ −4x2 x+1=−8−6x2+12x$\Rightarrow -4x^{2}x+1=-8-6x^2+12x$ 

   ⇒ 2x2−13x+9=0$\Rightarrow 2x^2-13x+9=0$ 

Here, degree of Equation is 2$2$.

Therefore, it is a Quadratic Equation.

2. Represent the following situations in the form of Quadratic Equations:

i). The area of the rectangular plot is 528 m2$m^2$. The length of the plot (in meters) is one more than twice its breadth. We need to find the length and breadth of the plot.

Ans:

We are given that area of a rectangular plot is 528m2$528m^2$ 

Let the breadth of the rectangular plot be x$x$ meters

Length is one more than twice its breadth

Therefore, length of rectangular plot is (2x+1)$\left(2x+1\right)$meters

Area of rectangle=$=$length ×$\times$ breadth

⇒ 528=x(2x+1)$\Rightarrow 528=x\left(2x+1\right)$ 

⇒ 528=2x2+x$\Rightarrow 528=2x^2+x$ 

⇒ 2x2+x 528=0$\Rightarrow 2x^2+x528=0$ 

This is a Quadratic Equation.

ii). The product of two consecutive numbers is 306. We need to find the integers.

Ans:

Let two consecutive numbers be xand(x+1).$x\text{and}\left(x+1\right).$ 

It is given that x(x+1) = 306 

⇒ x2+x=306$\Rightarrow x^2+x=306$ 

⇒ x2+x 306=0$\Rightarrow x^2+x306=0$ 

This is a Quadratic Equation.

iii). Rohan's mother is 26 years older than him. The product of their ages (in years) after 3 years will be 360. We would like to find Rohan's present age.

Ans:

Let present age of Rohan = x$=x$years

Let present age of Rohan's mother = (x +26)$=\left(x+26\right)$ years

Age of Rohan after 3$3$years = (x+3)$=\left(x+3\right)$ years 

Age of Rohan's mother after  3$3$ years = x+26+3 = (x+29)$=x+26+3=\left(x+29\right)$ years 

According to given condition: 

(x+3)(x+29)=360$\left(x+3\right)\left(x+29\right)=360$  

⇒ x2+29x+3x+87=360$\Rightarrow x^2+29x+3x+87=360$  

⇒ x2+32x 273=0$\Rightarrow x^2+32x273=0$ 

This is a Quadratic Equation.

iv). A train travels a distance of 480 km at a uniform speed. If the speed had been 8km/h less, then it would have taken 3 hours more to cover the same distance. We need to find the speed of the train.

Ans: 

Let the speed of the train be x$x$ km/h 

Time taken by train to cover 480 km = 480x$=480x$hours

If, speed had been 8km/h less than time taken would be (480x−8)$\left(480x-8\right)$  hours. According to given condition, if speed had been 8$8$km/h less than time taken is 3$3$hours less.

             Therefore, 480x 8=480x+3$480x8=480x+3$ 

             ⇒ 480(1x 8−1x)=3$\Rightarrow 480\left(1x8-1x\right)=3$ 

             ⇒ 480(x x+8) (x) (x−8)=3$\Rightarrow 480\left(xx+8\right)\left(x\right)\left(x-8\right)=3$ 

              ⇒ 480×8=3(x)(x−8)$\Rightarrow 480\times 8=3\left(x\right)\left(x-8\right)$ 

               ⇒ 3840=3x2−24x$\Rightarrow 3840=3x^2-24x$ 

     Dividing equation by3$3$, we get 

     ⇒ x2−8x 1280=0$\Rightarrow x^2-8x1280=0$ 

     This is a Quadratic Equation.

3. Find the roots of the following Quadratic Equations by factorization.

i). x2−3x 10=0$x^2-3x10=0$ 

Ans:

x2−3x 10=0$x^2-3x10=0$ 

⇒ x2−5x+2x 10=0$\Rightarrow x^2-5x+2x10=0$ 

      ⇒ x(x−5)+2(x−5)=0$\Rightarrow x\left(x-5\right)+2\left(x-5\right)=0$ 

      ⇒ (x−5)(x+2)=0$\Rightarrow \left(x-5\right)\left(x+2\right)=0$ 

       ⇒ x=5,−2$\Rightarrow x=5,-2$ 

ii). 2x2+x 6=0$2x^2+x6=0$ 

Ans:

2x2+x 6=0$2x^2+x6=0$ 

⇒ 2x2+4x−3x 6=0$\Rightarrow 2x^2+4x-3x6=0$

⇒ 2x(x+2)  3(x+2)=0$\Rightarrow 2x\left(x+2\right)3\left(x+2\right)=0$

⇒ (2x−3)(x+2)=0$\Rightarrow \left(2x-3\right)\left(x+2\right)=0$

⇒x=32,2$\Rightarrow x={\displaystyle \frac{3}{2}},2$ 

iii). 2–√x2+7x+52–√=0$\sqrt{2}{x}^2+7x+5\sqrt{2}=0$ 

Ans:

2–√x2+7x+52–√=0$\sqrt{2}{x}^2+7x+5\sqrt{2}=0$ 

          ⇒2–√x2+2x+5x+52–√=0$\Rightarrow \sqrt{2}{x}^2+2x+5x+5\sqrt{2}=0$ 

          ⇒2–√x2(x+2–√)+5(x+2–√)=0$\Rightarrow \sqrt{2}{x}^2\left(x+\sqrt{2}\right)+5\left(x+\sqrt{2}\right)=0$ 

          ⇒(2–√x+5)(x+2–√)=0$\Rightarrow \left(\sqrt{2}x+5\right)\left(x+\sqrt{2}\right)=0$ 

         ⇒x=−52–√,−2–√$\Rightarrow x={\displaystyle \frac{-5}{\sqrt{2}}},-\sqrt{2}$ 

           ⇒x=−52–√×2–√2–√,−2–√$\Rightarrow x={\displaystyle \frac{-5}{\sqrt{2}}}\times {\displaystyle \frac{\sqrt{2}}{\sqrt{2}}},-\sqrt{2}$

             ⇒x=−52–√2,−2–√$\Rightarrow x={\displaystyle \frac{-5\sqrt{2}}{2}},-\sqrt{2}$ 

iv). 2x2−x+18=0$2x^2-x+{\displaystyle \frac{1}{8}}=0$ 

Ans:

2x2−x+18=0$2x^2-x+{\displaystyle \frac{1}{8}}=0$ 

             ⇒16x2−8x+18=0$\Rightarrow {\displaystyle \frac{16x^2-8x+1}{8}}=0$ 

            ⇒ 16x2−8x+1=0$\Rightarrow 16x^2-8x+1=0$ 

          ⇒ 16x2−4x−4x+1=0$\Rightarrow 16x^2-4x-4x+1=0$ 

          ⇒ 4x(4x−1)  1(4x−1)=0$\Rightarrow 4x\left(4x-1\right)1\left(4x-1\right)=0$ 

             ⇒ (4x−1)(4x−1)=0$\Rightarrow \left(4x-1\right)\left(4x-1\right)=0$ 

            ⇒ x= 14,14$\Rightarrow x={\displaystyle \frac{1}{4}},{\displaystyle \frac{1}{4}}$ 

v). 100x2−20x+1=0$100x^2-20x+1=0$ 

Ans:

100x2−20x+1=0$100x^2-20x+1=0$ 

⇒ 100x2−10x−10x+1=0$\Rightarrow 100x^2-10x-10x+1=0$ 

⇒ 10x(10x−1)  1(10x−1)=0$\Rightarrow 10x\left(10x-1\right)1\left(10x-1\right)=0$ 

⇒ (10x−1)(10x−1)=0$\Rightarrow \left(10x-1\right)\left(10x-1\right)=0$ 

∴x=110,110$\therefore x={\displaystyle \frac{1}{10}},{\displaystyle \frac{1}{10}}$ 

4. Find the roots of the following equations:

i). x−1x=3,x≠0${\displaystyle \frac{x-1}{x}}=3,x\ne 0$ 

Ans:

x−1x=3wherex≠0$x-{\displaystyle \frac{1}{x}}=3wherex\ne 0$ 

⇒x2−1x=3$\Rightarrow {\displaystyle \frac{x^2-1}{x}}=3$ 

⇒ x 2 1=3x$\Rightarrow x{}^{2}1=3x$ 

⇒ x2−3x 1=0$\Rightarrow x^2-3x1=0$ 

Comparing equation x2−3x 1=0$x^2-3x1=0$with general formax2+bx+c=0$a{x}^2+bx+c=0$,

We get a=1,b=−3 and c=−1$a=1,b=-3andc=-1$ 

Using quadratic formulax=−b±b2−4ac−−−−−−−√2a$x={\displaystyle \frac{-b\pm \sqrt{b^2-4ac}}{2a}}$ to solve equation,

x=3±(3)2−4(1)(−1)−−−−−−−−−−−−−√2×1$x={\displaystyle \frac{3\pm \sqrt{{\left(3\right)}^2-4\left(1\right)\left(-1\right)}}{2\times 1}}$ 

             ⇒x=3±13−−√2$\Rightarrow x={\displaystyle \frac{3\pm \sqrt{13}}{2}}$ 

              ⇒x=3+13−−√2,3−13−−√2$\Rightarrow x={\displaystyle \frac{3+\sqrt{13}}{2}},{\displaystyle \frac{3-\sqrt{13}}{2}}$ 

(ii). 1x+4−1x−7=1130,x≠−4,7${\displaystyle \frac{1}{x+4}}-{\displaystyle \frac{1}{x-7}}={\displaystyle \frac{11}{30}},x\ne -4,7$ 

Ans:

1x+4−1x−7=1130wherex≠−4,7${\displaystyle \frac{1}{x+4}}-{\displaystyle \frac{1}{x-7}}={\displaystyle \frac{11}{30}}wherex\ne -4,7$

⇒ −30=x2−7x+4x28$\Rightarrow -30=x^2-7x+4x28$

⇒ x2−3x+2=0$\Rightarrow x^2-3x+2=0$ 

               Comparing equation x2−3x+2=0$x^2-3x+2=0$ with general formax2+bx+c=0$a{x}^2+bx+c=0$,

               We get a=1,b=−3 and c=2$a=1,b=-3andc=2$ 

               Using quadratic formula x=−b±b2−4ac−−−−−−−√2a$x={\displaystyle \frac{-b\pm \sqrt{b^2-4ac}}{2a}}$ to solve equation,

              x=3±(3)2−4(1)(2)−−−−−−−−−−−−√2×1$x={\displaystyle \frac{3\pm \sqrt{{\left(3\right)}^2-4\left(1\right)\left(2\right)}}{2\times 1}}$ 

              ⇒x=3±1–√2$\Rightarrow x={\displaystyle \frac{3\pm \sqrt{1}}{2}}$ 

              ⇒x=3+1–√2,3−1–√2$\Rightarrow x={\displaystyle \frac{3+\sqrt{1}}{2}},{\displaystyle \frac{3-\sqrt{1}}{2}}$ 

              ⇒x=2,1$\Rightarrow x=2,1$ 

5. The sum of reciprocals of Rehman's ages (in years) 3$3$  years ago and 5$5$  years from now is 13$13$. Find his present age.

Ans:

Let present age of Rehman= x$=x$  years 

Age of Rehman 3$3$ years ago = (x−3)$=\left(x-3\right)$  years.

 Age of Rehman after 5$5$  years = (x+5)$=\left(x+5\right)$ years

 According to the given condition:

1x−3+1x+5=13${\displaystyle \frac{1}{x-3}}+{\displaystyle \frac{1}{x+5}}={\displaystyle \frac{1}{3}}$ 

⇒(x+5)+(x−3)(x−3)(x+5)=13$\Rightarrow {\displaystyle \frac{\left(x+5\right)+\left(x-3\right)}{\left(x-3\right)\left(x+5\right)}}={\displaystyle \frac{1}{3}}$ 

⇒ 3(2x+2) =(x−3)(x+5) $\Rightarrow 3\left(2x+2\right)=\left(x-3\right)\left(x+5\right)$ 

⇒ 6x+6=x2−3x+5x−15$\Rightarrow 6x+6=x^2-3x+5x-15$ 

 ⇒ x2−4x 15  6=0$\Rightarrow x^2-4x156=0$ 

⇒x2−4x 21=0$\Rightarrow x^2-4x21=0$ 

Comparing quadratic equation x x2−4x 21=0$x^2-4x21=0$ with general form ax2+bx+c=0$a{x}^2+bx+c=0$, We get a=1,b=−4 and c=−21$a=1,b=-4andc=-21$ 

Using quadratic formulax=−b±b2−4ac−−−−−−−√2a$x={\displaystyle \frac{-b\pm \sqrt{b^2-4ac}}{2a}}$

x=4±(4)2−4(1)(−21)−−−−−−−−−−−−−−√2×1$x={\displaystyle \frac{4\pm \sqrt{{\left(4\right)}^2-4\left(1\right)\left(-21\right)}}{2\times 1}}$ 

⇒x=4±16+84−−−−−−√2$\Rightarrow x={\displaystyle \frac{4\pm \sqrt{16+84}}{2}}$ 

⇒x=4±100−−−√2=4±102$\Rightarrow x={\displaystyle \frac{4\pm \sqrt{100}}{2}}={\displaystyle \frac{4\pm 10}{2}}$ 

⇒x=4+102,4−102$\Rightarrow x={\displaystyle \frac{4+10}{2}},{\displaystyle \frac{4-10}{2}}$ 

∴x=7,−3$\therefore x=7,-3$ 

We discard x=−3$x=-3$ .Since age cannot be in negative.

Therefore, present age of Rehman is 7$7$ years

6. Two water taps together can fill a tank in 938$9{\displaystyle \frac{3}{8}}$hours. The tap of larger diameter takes 10$10$ hours less than the smaller one to fill the tank separately. Find the time in which each tap can separately fill the tank.

Ans: 

Let time taken by tap of smaller diameter to fill the tank = x$=x$ hours 

Let time taken by tap of larger diameter to fill the tank = (x 10)$=\left(x10\right)$ hours 

It means that tap of smaller diameter fills 1xth${{\displaystyle \frac{1}{x}}}^{th}$ part of tank in 1$1$  hour. … (1)

 And, tap of larger diameter fills 1x−10th${{\displaystyle \frac{1}{x-10}}}^{th}$ part of tank in 1$1$  hour. … (2) 

When two taps are used together, they fill tank in 758$758$ hours

In 1 hour, they fill875th${{\displaystyle \frac{8}{75}}}^{th}$ part of tank ⎡⎣⎢⎢1758=875⎤⎦⎥⎥$\left[{\displaystyle \frac{1}{{\displaystyle \frac{75}{8}}}}={\displaystyle \frac{8}{75}}\right]$ … (3)

From (1), (2) and (3),

1x+1x−10=875${\displaystyle \frac{1}{x}}+{\displaystyle \frac{1}{x-10}}={\displaystyle \frac{8}{75}}$ 

⇒x−10+xx(x−10)=875$\Rightarrow {\displaystyle \frac{x-10+x}{x\left(x-10\right)}}={\displaystyle \frac{8}{75}}$ 

⇒ 75(2x−10)=8(x2−10x)$\Rightarrow 75\left(2x-10\right)=8\left(x^2-10x\right)$ 

⇒ 150x 750=8x2−80x$\Rightarrow 150x750=8x^2-80x$ 

⇒ 8x2 −80x−150x+750=0$\Rightarrow 8x^2-80x-150x+750=0$ 

Comparing equation 4x2 −115x+375=0$4x^2-115x+375=0$ with general equationax2+bx+c=0$a{x}^2+bx+c=0$, 

We get a=4,b=−115andc=375$a=4,b=-115andc=375$ 

Applying quadratic formula x=−b±b2−4ac−−−−−−−√2a$x={\displaystyle \frac{-b\pm \sqrt{b^2-4ac}}{2a}}$

x=115±(−115)2−4(4)(375)−−−−−−−−−−−−−−−−−√2×4$x={\displaystyle \frac{115\pm \sqrt{{\left(-115\right)}^2-4\left(4\right)\left(375\right)}}{2\times 4}}$ 

⇒x=115±13225−6000−−−−−−−−−−−√8$\Rightarrow x={\displaystyle \frac{115\pm \sqrt{13225-6000}}{8}}$ 

⇒115±7225−−−−√8$\Rightarrow {\displaystyle \frac{115\pm \sqrt{7225}}{8}}$ 

⇒115+858,115−858$\Rightarrow {\displaystyle \frac{115+85}{8}},{\displaystyle \frac{115-85}{8}}$ 

∴x=25,3.75$\therefore x=25,3.75$ 

Time taken by larger tap = x 10=3.75  10=−6.25$=x10=3.7510=-6.25$ hours 

Time cannot be in negative. Therefore, we ignore this value. 

Time taken by larger tap = x 10=25  10=15$=x10=2510=15$ hours 

Therefore, time taken by larger tap is 15$15$ hours and time taken by smaller tap is 25$25$  hours.

7. Find the nature of the roots of the following quadratic equations. If the real roots exist, find them.

i). 2x2 3x + 5 = 0$2x^{2}3x+5=0$

Ans:

2x2 3x + 5 = 0$2x^{2}3x+5=0$

Comparing this equation with general equationax2+bx+c=0$a{x}^2+bx+c=0$ , 

We get a=2,b=−3 and c=5$a=2,b=-3andc=5$ 

Discriminant= b2−4ac=(−3)2 4(2)(5)=9  40=−31$=b^2-4ac={\left(-3\right)}^{2}4\left(2\right)\left(5\right)=940=-31$ 

Discriminant is less than 0 which means the equation has no real roots.

ii). 3x2−43–√x+4=0$3x^2-4\sqrt{3}x+4=0$ 

Ans:

3x2−43–√x+4=0$3x^2-4\sqrt{3}x+4=0$

Comparing this equation with general equation ax2+bx+c=0$a{x}^2+bx+c=0$, 

We get a=3,b=−43–√ and c=4$a=3,b=-4\sqrt{3}andc=4$

Discriminant= b2−4ac=(−43–√)2−4(3)(4)=48  48=0$=b^2-4ac={\left(-4\sqrt{3}\right)}^2-4\left(3\right)\left(4\right)=4848=0$ 

Discriminant is equal to zero which means equations have equal real roots. Applying quadraticx=−b±b2−4ac−−−−−−−√2a$x={\displaystyle \frac{-b\pm \sqrt{b^2-4ac}}{2a}}$ to find roots,

             x=43–√±0–√6=23–√3$x={\displaystyle \frac{4\sqrt{3}\pm \sqrt{0}}{6}}={\displaystyle \frac{2\sqrt{3}}{3}}$ 

 Because, equation has two equal roots, it means x=23–√3,23–√3$x={\displaystyle \frac{2\sqrt{3}}{3}},{\displaystyle \frac{2\sqrt{3}}{3}}$ 

iii). 2x2 +6x + 3 = 0$2x^2+6x+3=0$ 

Ans:

2x2 +6x + 3 = 0$2x^2+6x+3=0$

         Comparing equation with general equationax2+bx+c=0$a{x}^2+bx+c=0$ ,

          We get a=2,b=−6, and c=3$a=2,b=-6,andc=3$ 

          Discriminant = b2−4ac=(−6)2− 4(2)(3)=36 − 24=12$=b^2-4ac={\left(-6\right)}^2-4\left(2\right)\left(3\right)=36-24=12$ 

          Value of the discriminant is greater than zero. 

          Therefore, the equation has distinct and real roots.

         Applying quadratic formula x=−b±b2−4ac−−−−−−−√2a$x={\displaystyle \frac{-b\pm \sqrt{b^2-4ac}}{2a}}$ to find roots,

          x=6±12−−√4=6±23–√4$x={\displaystyle \frac{6\pm \sqrt{12}}{4}}={\displaystyle \frac{6\pm 2\sqrt{3}}{4}}$ 

          ⇒x=3±3–√2$\Rightarrow x={\displaystyle \frac{3\pm \sqrt{3}}{2}}$ 

          ⇒x=3+3–√2,3−3–√2$\Rightarrow x={\displaystyle \frac{3+\sqrt{3}}{2}},{\displaystyle \frac{3-\sqrt{3}}{2}}$ 

8. If −4$-4$ is a root of the quadratic equation and the quadratic equationx2+px−4$x^2+px-4$ has equal root, find the value of k.$k.$ 

Ans: 

−4$-4$is root of x2+px−4=0$x^2+px-4=0$

∴(−4)2+p(−4)−4=0$\therefore {\left(-4\right)}^2+p\left(-4\right)-4=0$ 

⇒16−4p−4=0$\Rightarrow 16-4p-4=0$ 

⇒−4p=−12$\Rightarrow -4p=-12$ 

⇒p=3$\Rightarrow p=3$ 

x2+px+k=0$x^2+px+k=0$ (Given)

x2+3x+k=0$x^2+3x+k=0$ 

D=b2−4ac$D=b^2-4ac$ 

⇒0=(3)2−4×1×k$\Rightarrow 0={\left(3\right)}^2-4\times 1\times k$ ForequalrootsD=0$Fo\mathrm{r }equa\mathrm{l }root\mathrm{s }D=0$

⇒4k=9$\Rightarrow 4k=9$ 

⇒k=94$\Rightarrow k={\displaystyle \frac{9}{4}}$ 

9. Solve for x:5x+1+51−x=26$x:5^{x+1}+5^{1-x}=26$ 

Ans: 

5x+1+51−x=26$5^{x+1}+5^{1-x}=26$ 

     5x.51+51.5−x=26$5^x{.5}^1+5^1{.5}^{-x}=26$ 

     ⇒5x.5+515x=26$\Rightarrow 5^x.5+{\displaystyle \frac{5^1}{5^x}}=26$ 

     Put 5x=y$5^x=y$ 

      5y1+5y=26${\displaystyle \frac{5y}{1}}+{\displaystyle \frac{5}{y}}=26$ 

      ⇒5y2−26y+5=0$\Rightarrow 5y^2-26y+5=0$ 

      ⇒5y2−25y−y+5=0$\Rightarrow 5y^2-25y-y+5=0$ 

      ⇒5y(y−5)−1(y−5)=0$\Rightarrow 5y\left(y-5\right)-1\left(y-5\right)=0$ 

      ⇒$\Rightarrow$ (y−5)(5y−1)=0$\left(y-5\right)\left(5y-1\right)=0$ 

      ⇒y=5$\Rightarrow y=5$ or  y=15$y={\displaystyle \frac{1}{5}}$ 

      But 

      5x=51$5^x=5^1$ and 5x=15$5^x={\displaystyle \frac{1}{5}}$ 

      ⇒x=1$\Rightarrow x=1$ and 5x=5−1⇒x=−1$5^x=5^{-1}\Rightarrow x=-1$ 

10. 1p+q+x=1p+1q+1x${\displaystyle \frac{1}{p+q+x}}={\displaystyle \frac{1}{p}}+{\displaystyle \frac{1}{q}}+{\displaystyle \frac{1}{x}}$ solve forx$x$ by factorization method.

Ans:

1p+q+x=1p+1q+1x${\displaystyle \frac{1}{p+q+x}}={\displaystyle \frac{1}{p}}+{\displaystyle \frac{1}{q}}+{\displaystyle \frac{1}{x}}$ 

⇒1p+q+x−1x=1p+1q$\Rightarrow {\displaystyle \frac{1}{p+q+x}}-{\displaystyle \frac{1}{x}}={\displaystyle \frac{1}{p}}+{\displaystyle \frac{1}{q}}$

⇒x−p−q−xx2+px+qx=p+qpq$\Rightarrow {\displaystyle \frac{x-p-q-x}{x^2+px+qx}}={\displaystyle \frac{p+q}{pq}}$ 

⇒−(p+q)x2+px+qx=p+qpq$\Rightarrow {\displaystyle \frac{-\left(p+q\right)}{x^2+px+qx}}={\displaystyle \frac{p+q}{pq}}$ 

⇒−1x2+px+qx=1pq$\Rightarrow {\displaystyle \frac{-1}{x^2+px+qx}}={\displaystyle \frac{1}{pq}}$ 

⇒x2+px+qx=−pq$\Rightarrow x^2+px+qx=-pq$ 

⇒x2+px+qx+pq=0$\Rightarrow x^2+px+qx+pq=0$ 

⇒x(x+p)+q(x+p)=0$\Rightarrow x\left(x+p\right)+q\left(x+p\right)=0$ 

⇒(x+p)(x+q)=0$\Rightarrow \left(x+p\right)\left(x+q\right)=0$ 

∴x=−porx=−q$\therefore x=-porx=-q$ 

11. 5x2−6x−2=0$5x^2-6x-2=0$, solve forx$x$  by the method of completing the square.

Ans: 

5x2−6x−2=0$5x^2-6x-2=0$

⇒x2−65x−25=0$\Rightarrow x^2-{\displaystyle \frac{6}{5}}x-{\displaystyle \frac{2}{5}}=0$ 

⇒x2−65x+(35)2−25=0$\Rightarrow x^2-{\displaystyle \frac{6}{5}}x+{\left({\displaystyle \frac{3}{5}}\right)}^2-{\displaystyle \frac{2}{5}}=0$ 

⇒(x−35)2=925+25$\Rightarrow {\left(x-{\displaystyle \frac{3}{5}}\right)}^2={\displaystyle \frac{9}{25}}+{\displaystyle \frac{2}{5}}$ 

⇒(x−35)2=9+1025$\Rightarrow {\left(x-{\displaystyle \frac{3}{5}}\right)}^2={\displaystyle \frac{9+10}{25}}$ 

⇒(x−35)2=1925$\Rightarrow {\left(x-{\displaystyle \frac{3}{5}}\right)}^2={\displaystyle \frac{19}{25}}$ 

⇒x−35=±19−−√5$\Rightarrow x-{\displaystyle \frac{3}{5}}=\pm {\displaystyle \frac{\sqrt{19}}{5}}$ 

⇒x=35±19−−√5$\Rightarrow x={\displaystyle \frac{3}{5}}\pm {\displaystyle \frac{\sqrt{19}}{5}}$ 

⇒x=3+19−−√5orx=3−19−−√5$\Rightarrow x={\displaystyle \frac{3+\sqrt{19}}{5}}orx={\displaystyle \frac{3-\sqrt{19}}{5}}$ 

12. Solve forx:a2b2x2+b2x−a2x−1=0$x:a^2{b}^2{x}^2+b^{2}x-a^{2}x-1=0$ 

Ans:

a2b2x2+b2x−a2x−1=0$a^2{b}^2{x}^2+b^{2}x-a^{2}x-1=0$

⇒b2x(a2x+1)−1(a2x−1)=0$\Rightarrow b^{2}x\left(a^{2}x+1\right)-1\left(a^{2}x-1\right)=0$ 

⇒(a2x+1)(b2x−1)=0$\Rightarrow \left(a^{2}x+1\right)\left(b^{2}x-1\right)=0$ 

∴x=−1a2orx=−1b2$\therefore x={\displaystyle \frac{-1}{a^2}}orx={\displaystyle \frac{-1}{b^2}}$ 

13. Using quadratic formula, solve for x:9x2−9(a+b)x+(2a2+5ab+2b2)=0$x:9x^2-9\left(a+b\right)x+\left(2a^2+5ab+2b^2\right)=0$ 

Ans:

D=b2−4ac$D=b^2-4ac$ 

=(−9(a+b))2−4×9×(2a2+5ab+2ab2)$={\left(-9\left(a+b\right)\right)}^2-4\times 9\times \left(2a^2+5ab+2a{b}^2\right)$ 

=81(a+b)2−36(2a2+5ab+2b2)$=81{\left(a+b\right)}^2-36\left(2a^2+5ab+2b^2\right)$ 

=9[9(a2+b2+2ab−8a2−20ab−8b2)]$=9\left[9\left(a^2+b^2+2ab-8a^2-20ab-8b^2\right)\right]$ 

=9[a2+b2−2ab]$=9\left[a^2+b^2-2ab\right]$ 

=9(a−b)2$=9{\left(a-b\right)}^2$

x=−b±D−−√2a=9(a+b)±9(a−b)2−−−−−−−√2×9$x={\displaystyle \frac{-b\pm \sqrt{D}}{2a}}={\displaystyle \frac{9\left(a+b\right)\pm \sqrt{9{\left(a-b\right)}^2}}{2\times 9}}$ 

⇒x=3[3(a+b)±(a−b)]2×9$\Rightarrow x=3{\displaystyle \frac{\left[3\left(a+b\right)\pm \left(a-b\right)\right]}{2\times 9}}$ 

⇒x=(3a+3b)±(a−b)6$\Rightarrow x={\displaystyle \frac{\left(3a+3b\right)\pm \left(a-b\right)}{6}}$ 

⇒x=3a+3b+a−b6orx=3a+3b+a−b6$\Rightarrow x={\displaystyle \frac{3a+3b+a-b}{6}}orx={\displaystyle \frac{3a+3b+a-b}{6}}$ 

⇒x=4a+2b6orx=4a+2b6$\Rightarrow x={\displaystyle \frac{4a+2b}{6}}orx={\displaystyle \frac{4a+2b}{6}}$ 

⇒x=2a+b3orx=2a+b3$\Rightarrow x={\displaystyle \frac{2a+b}{3}}orx={\displaystyle \frac{2a+b}{3}}$ 

14. In a cricket match, Kapil took one wicket less than twice the number of wickets taken by Ravi. If the product of the numbers of wickets taken by these two is 15, find the number of wickets taken by each.

Ans: 

Let no. of wicket taken by Ravi =x$=x$

 of wicket taken by Kapil =2x−1$=2x-1$ 

According to question,

(2x−1)x=15$\left(2x-1\right)x=15$ 

⇒2x2−x−15=0$\Rightarrow 2x^2-x-15=0$ 

∴x=3orx=−52$\therefore x=3orx={\displaystyle \frac{-5}{2}}$ (Neglects)

So, no. of wickets taken by Ravi is  x=3$x=3$ 

 

15. The sum of a number and its reciprocal is 174${\displaystyle \frac{17}{4}}$ . Find the number.

Ans: 

Let no. be x$x$ 

According to question,

x1+1x=174${\displaystyle \frac{x}{1}}+{\displaystyle \frac{1}{x}}={\displaystyle \frac{17}{4}}$ 

⇒x2+1x=174$\Rightarrow {\displaystyle \frac{x^2+1}{x}}={\displaystyle \frac{17}{4}}$ 

⇒4x2+4=17x$\Rightarrow 4x^2+4=17x$ 

⇒4x2−17x+4=0$\Rightarrow 4x^2-17x+4=0$ 

⇒4x2−16x−x+4=0$\Rightarrow 4x^2-16x-x+4=0$ 

⇒4x(x−4)−1(x−4)=0$\Rightarrow 4x\left(x-4\right)-1\left(x-4\right)=0$ 

⇒(x−4)(4x−1)=0$\Rightarrow \left(x-4\right)\left(4x-1\right)=0$ 

∴x=4orx=14$\therefore x=4orx={\displaystyle \frac{1}{4}}$ 

 

4 Marks Questions

1. Find the roots of the following Quadratic Equations by applying quadratic formulas.

i). 2x2− 7x+ 3 = 0$2x^2-7x+3=0$

Ans:

2x2− 7x+ 3 = 0$2x^2-7x+3=0$

Comparing quadratic equation 2x2− 7x+ 3 = 0$2x^2-7x+3=0$with general form ax2+bx+c=0$a{x}^2+bx+c=0$, we get a=2, b=−7 and c=3$a=2,b=-7andc=3$ 

 Putting these values in quadratic formulax=−b±b2−4ac−−−−−−−√2a$x={\displaystyle \frac{-b\pm \sqrt{b^2-4ac}}{2a}}$

x=7±(−7)2−4(2)(3)−−−−−−−−−−−−−√2×2$x={\displaystyle \frac{7\pm \sqrt{{\left(-7\right)}^2-4\left(2\right)\left(3\right)}}{2\times 2}}$ 

⇒x=7±49−24−−−−−−√4$\Rightarrow x={\displaystyle \frac{7\pm \sqrt{49-24}}{4}}$ 

⇒x=7±54$\Rightarrow x={\displaystyle \frac{7\pm 5}{4}}$ 

⇒x=7+54,7−54$\Rightarrow x={\displaystyle \frac{7+5}{4}},{\displaystyle \frac{7-5}{4}}$ 

∴x=3,12$\therefore x=3,{\displaystyle \frac{1}{2}}$ 

ii). 2x2+ x  4 = 0$2x^2+x4=0$ 

Ans:

Comparing quadratic equation 2x2+ x  4 = 0$2x^2+x4=0$with the general formax2+bx+c=0$a{x}^2+bx+c=0$, we get a=2, b=1 and c=−4$a=2,b=1andc=-4$ 

Putting these values in quadratic formulax=−b±b2−4ac−−−−−−−√2a$x={\displaystyle \frac{-b\pm \sqrt{b^2-4ac}}{2a}}$

x=−1±12−4(2)(−4)−−−−−−−−−−−−√2×2$x={\displaystyle \frac{-1\pm \sqrt{1^2-4\left(2\right)\left(-4\right)}}{2\times 2}}$ 

⇒x=−1±33−−√4$\Rightarrow x={\displaystyle \frac{-1\pm \sqrt{33}}{4}}$ 

⇒x=−1−33−−√4,−1+33−−√4$\Rightarrow x={\displaystyle \frac{-1-\sqrt{33}}{4}},{\displaystyle \frac{-1+\sqrt{33}}{4}}$ 

iii). 4x2+43–√x+3=0$4x^2+4\sqrt{3}x+3=0$ 

Ans:

Comparing quadratic equation 4x2+43–√x+3=0$4x^2+4\sqrt{3}x+3=0$ with the general formax2+bx+c=0$a{x}^2+bx+c=0$, we get a=4, b=43–√ and c=3$a=4,b=4\sqrt{3}andc=3$ 

x=−43–√±(43–√)2−4(4)(3)−−−−−−−−−−−−−−√2×4$x={\displaystyle \frac{-4\sqrt{3}\pm \sqrt{{\left(4\sqrt{3}\right)}^2-4\left(4\right)\left(3\right)}}{2\times 4}}$ 

⇒x=−43–√±0–√8$\Rightarrow x={\displaystyle \frac{-4\sqrt{3}\pm \sqrt{0}}{8}}$ 

⇒x=−3–√2$\Rightarrow x={\displaystyle \frac{-\sqrt{3}}{2}}$ 

A quadratic equation has two roots. Here, both the roots are equal.

Therefore,x=−3–√2,−3–√2$x={\displaystyle \frac{-\sqrt{3}}{2}},{\displaystyle \frac{-\sqrt{3}}{2}}$ 

iv). 2x2+ x + 4 = 0$2x^2+x+4=0$ 

Ans:

2x2+ x + 4 = 0$2x^2+x+4=0$

Comparing quadratic equation 2x2+ x + 4 = 0$2x^2+x+4=0$ with the general formax2+bx+c=0$a{x}^2+bx+c=0$, we get a=2,b=1 and c= 4$a=2,b=1andc=4$ 

Putting these values in quadratic formulax=−b±b2−4ac−−−−−−−√2a$x={\displaystyle \frac{-b\pm \sqrt{b^2-4ac}}{2a}}$

x=−1±(1)2−4(2)(4)−−−−−−−−−−−−√2×2$x={\displaystyle \frac{-1\pm \sqrt{{\left(1\right)}^2-4\left(2\right)\left(4\right)}}{2\times 2}}$ 

⇒x=−1±−31−−−−√4$\Rightarrow x={\displaystyle \frac{-1\pm \sqrt{-31}}{4}}$ 

But, the square root of a negative numbers is not defined. 

Therefore, Quadratic Equation 2x2+ x + 4 = 0$2x^2+x+4=0$ has no solution.

2. An express train takes 1$1$ hour less than a passenger train to travel 132$132$  km between Mysore and Bangalore (without taking into consideration the time they stop at intermediate stations). If, the average speed of the express train is 11$11$  km/h more than that of the passenger train, find the average speed of two trains

Ans: Let average speed of passenger train = x$=x$ km/h 

Let average speed of express train = (x+11)$=\left(x+11\right)$ km/h 

Time taken by passenger train to cover 132$132$ km =$=$ 132x${\displaystyle \frac{132}{x}}$  hours 

Time taken by express train to cover132$132$km =(132x+11)$=\left({\displaystyle \frac{132}{x+11}}\right)$  hours

According to the given condition

132x=132x+11+1${\displaystyle \frac{132}{x}}={\displaystyle \frac{132}{x+11}}+1$ 

⇒132(1x−1x+11)=1$\Rightarrow 132\left({\displaystyle \frac{1}{x}}-{\displaystyle \frac{1}{x+11}}\right)=1$ 

132(x+11−xx(+11))=1$132\left({\displaystyle \frac{x+11-x}{x\left(+11\right)}}\right)=1$ 

⇒ 132(11)=x(x+11)$\Rightarrow 132\left(11\right)=x\left(x+11\right)$ 

⇒ 1452=x2+11x$\Rightarrow 1452=x^2+11x$ 

⇒ x2+11x 1452=0$\Rightarrow x^2+11x1452=0$ 

Comparing equation x2+11x 1452=0$x^2+11x1452=0$with general quadratic equationax2+bx+c=0$a{x}^2+bx+c=0$, we get a=1,b=11 and c=−1452$a=1,b=11andc=-1452$ 

Applying Quadratic Formulax=−b±b2−4ac−−−−−−−√2a$x={\displaystyle \frac{-b\pm \sqrt{b^2-4ac}}{2a}}$

x=−11±(11)2−4(1)(−1452)−−−−−−−−−−−−−−−−−√2×1$x={\displaystyle \frac{-11\pm \sqrt{{\left(11\right)}^2-4\left(1\right)\left(-1452\right)}}{2\times 1}}$ 

⇒x=−11±121+5808−−−−−−−−−√2$\Rightarrow x={\displaystyle \frac{-11\pm \sqrt{121+5808}}{2}}$ 

⇒x=−11±5929−−−−√2$\Rightarrow x={\displaystyle \frac{-11\pm \sqrt{5929}}{2}}$ 

⇒x=−11±772$\Rightarrow x={\displaystyle \frac{-11\pm 77}{2}}$ 

⇒x=−11+772,−11−772$\Rightarrow x={\displaystyle \frac{-11+77}{2}},{\displaystyle \frac{-11-77}{2}}$ 

∴x=33,−44$\therefore x=33,-44$ 

As speed cannot be negative. Therefore, speed of passenger train = 33$=33$ km/h

And, speed of express train = x+11=33+11=44$=x+11=33+11=44$ km/h

3. Sum of areas of two squares is 468 m2$468m^2$. If, the difference of their perimeters is 24$24$ meters, find the sides of the two squares.

Ans: 

Let perimeter of first square = x$=x$ meters

Let perimeter of second square = (x+24)$=\left(x+24\right)$ meters 

Length of side of first square =$=$x4${\displaystyle \frac{x}{4}}$  meters {Perimeter of square = 4 ×$=4\times$ length of side}

Length of side of second square =$=$ =(x+244)$=\left({\displaystyle \frac{x+24}{4}}\right)$  meters 

Area of first square =side×side$=side\times side$

=x4×x4=x216m2${\displaystyle \frac{x}{4}}\times {\displaystyle \frac{x}{4}}={\displaystyle \frac{x^2}{16}}{m}^2$ 

Area of second square =(x+244)2m2$={\left({\displaystyle \frac{x+24}{4}}\right)}^2{m}^2$ 

According to given condition:

x216+(x+244)2=468${\displaystyle \frac{x^2}{16}}+{\left({\displaystyle \frac{x+24}{4}}\right)}^2=468$ 

⇒x216+x2+576+48x16=468$\Rightarrow {\displaystyle \frac{x^2}{16}}+{\displaystyle \frac{x^2+576+48x}{16}}=468$ 

⇒x2+x2+576+48x16=468$\Rightarrow {\displaystyle \frac{x^2+x^2+576+48x}{16}}=468$ 

⇒ 2x2+576+48x=468×16$\Rightarrow 2x^2+576+48x=468\times 16$ 

⇒ 2x2+48x+576=7488 $\Rightarrow 2x^2+48x+576=7488$ 

⇒ 2x2+48x 6912=0 $\Rightarrow 2x^2+48x6912=0$ 

⇒ x2+24x 3456=0$\Rightarrow x^2+24x3456=0$ 

Comparing equationx2+24x 3456=0$x^2+24x3456=0$ with standard formax2+bx+c=0$a{x}^2+bx+c=0$, We get a=1,b=24 andc= −3456$a=1,b=24andc=-3456$

Applying Quadratic Formulax=−b±b2−4ac−−−−−−−√2a$x={\displaystyle \frac{-b\pm \sqrt{b^2-4ac}}{2a}}$

x=−24±(24)2−4(1)(−3456)−−−−−−−−−−−−−−−−−√2×1$x={\displaystyle \frac{-24\pm \sqrt{{\left(24\right)}^2-4\left(1\right)\left(-3456\right)}}{2\times 1}}$ 

⇒x=−24±576+13824−−−−−−−−−−√2$\Rightarrow x={\displaystyle \frac{-24\pm \sqrt{576+13824}}{2}}$ 

⇒x=−24±14400−−−−−√2=−24±1202$\Rightarrow x={\displaystyle \frac{-24\pm \sqrt{14400}}{2}}={\displaystyle \frac{-24\pm 120}{2}}$ 

⇒x=−24+1202,−24−1202$\Rightarrow x={\displaystyle \frac{-24+120}{2}},{\displaystyle \frac{-24-120}{2}}$ 

∴x=48,−72$\therefore x=48,-72$ 

Perimeter of square cannot be in negative. Therefore, we discardx=−72$x=-72$. 

Therefore, perimeter of first square = 48$=48$meters 

And, Perimeter of second square = x+24=48+24=72$=x+24=48+24=72$meters 

⇒$\Rightarrow$ Side of First square =Perimeter4=484=12m$={\displaystyle \frac{Perimeter}{4}}={\displaystyle \frac{48}{4}}=12m$  

And, Side of second Square =Perimeter4=724=18m$={\displaystyle \frac{Perimeter}{4}}={\displaystyle \frac{72}{4}}=18m$ 

4. Is it possible to design a rectangular park of perimeter 80$80$ meters and area400 m2$400m^2$. If so, find its length and breadth.

Ans: 

Let length of park = x$=x$ meters

 We are given area of rectangular park = 400 m2$=400m^2$ 

Therefore, breadth of park =$=$ 400x${\displaystyle \frac{400}{x}}$  meters {Area of rectangle =$=$ length ×$\times$ breadth}

Perimeter of rectangular park = 2$=2$ (length+$+$breath)=$=$ (x+400x)$\left(x+{\displaystyle \frac{400}{x}}\right)$ meters

We are given perimeter of rectangle = 80$=80$ meters

 According to condition:

2(x+400x)=80$2\left(x+{\displaystyle \frac{400}{x}}\right)=80$ 

⇒2(x2+400x)=80$\Rightarrow 2\left({\displaystyle \frac{x^2+400}{x}}\right)=80$ 

  ⇒ 2x2+800=80x$\Rightarrow 2x^2+800=80x$ 

⇒ 2x2−80x+800=0$\Rightarrow 2x^2-80x+800=0$ 

⇒ x2−40x+400=0$\Rightarrow x^2-40x+400=0$ 

Comparing equation, x2−40x+400=0$x^2-40x+400=0$ with general quadratic equationax2+bx+c=0$a{x}^2+bx+c=0$, we get a=1,b=−40 and c=400$a=1,b=-40andc=400$ 

Discriminant= b2−4ac=(−40)2 4(1)(400)=1600  1600=0$=b^2-4ac={\left(-40\right)}^{2}4\left(1\right)\left(400\right)=16001600=0$ 

Discriminant is equal to0$0$ .

Therefore, two roots of equation are real and equal which means that it is possible to design a rectangular park of perimeter 80$80$meters and area400 m2$400m^2$.

Using quadratic formulax=−b±b2−4ac−−−−−−−√2a$x={\displaystyle \frac{-b\pm \sqrt{b^2-4ac}}{2a}}$  to solve equation,s

x=40±0–√2=422=20$x={\displaystyle \frac{40\pm \sqrt{0}}{2}}={\displaystyle \frac{42}{2}}=20$ 

Here, both the roots are equal to20$20$.

 Therefore, length of rectangular park = 20$=20$meters

Breadth of rectangular park=400x=40020=20m$={\displaystyle \frac{400}{x}}={\displaystyle \frac{400}{20}}=20m$ 

5. If I had walked 1$1$  km per hour faster, I would have taken 10$10$  minutes less to walk 2$2$  km. Find the rate of my walking.

Ans: Distance = 2$=2$ km

 Let speed =x$=x$  km/hr.

 New speed = (x+1)$=\left(x+1\right)$ km/hr. 

Time taken by normal speed =2xhr$={\displaystyle \frac{2}{x}}hr$ 

Time taken by new speed = 2x+1hr${\displaystyle \frac{2}{x+1}}hr$ 

According to question,

2x−2x+1=1060${\displaystyle \frac{2}{x}}-{\displaystyle \frac{2}{x+1}}={\displaystyle \frac{10}{60}}$ 

⇒2x+2−2xx2+x=16$\Rightarrow {\displaystyle \frac{2x+2-2x}{x^2+x}}={\displaystyle \frac{1}{6}}$ 

⇒x2+x=12$\Rightarrow x^2+x=12$ 

⇒x2+x−12=0$\Rightarrow x^2+x-12=0$ 

⇒x2+4x−3x−12=0$\Rightarrow x^2+4x-3x-12=0$ 

⇒x(x+4)−3(x+4)=0$\Rightarrow x\left(x+4\right)-3\left(x+4\right)=0$ 

⇒(x+4)(x−3)=0$\Rightarrow \left(x+4\right)\left(x-3\right)=0$ 

∴x=−4orx=3$\therefore x=-4orx=3$ 

So, speed is x=3$x=3$ km/hr

6. A takes 6$6$  days less than the time taken by B to finish a piece of work. If both A and B together can finish it in 4$4$  days, find the time taken by B to finish the work.

Ans: Let B takes x$x$  days to finish the work, then A alone can finish it in (x−6)$\left(x-6\right)$ days 

According to question,

1x+1x−6=14${\displaystyle \frac{1}{x}}+{\displaystyle \frac{1}{x-6}}={\displaystyle \frac{1}{4}}$ 

⇒x−6+xx2−6x=14$\Rightarrow {\displaystyle \frac{x-6+x}{x^2-6x}}={\displaystyle \frac{1}{4}}$ 

⇒2x−6x2−6x=14$\Rightarrow {\displaystyle \frac{2x-6}{x^2-6x}}={\displaystyle \frac{1}{4}}$ 

⇒x2−6x=8x−24$\Rightarrow x^2-6x=8x-24$ 

⇒x2−14x+24=0$\Rightarrow x^2-14x+24=0$ 

⇒x2−12x−2x+24=0$\Rightarrow x^2-12x-2x+24=0$ 

=x(x−12)−2(x−12)=0$=x\left(x-12\right)-2\left(x-12\right)=0$ 

⇒(x−12)(x−2)=0$\Rightarrow \left(x-12\right)\left(x-2\right)=0$ 

∴x=12orx=2$\therefore x=12orx=2$ 

x=2$x=2$ (Neglect)

So, B takesx=12$x=12$ days.

7. A plane left 30$30$ minutes later than the schedule time and in order to reach its 59$59$  destination 1500$1500$  km away in time it has to increase its speed by 250$250$km/hr from its usual speed. Find its usual speed.

Ans: Let usual speed=x$=x$ km/hr 

New speed =(x+250)$=\left(x+250\right)$ km/hr

Total distance = 1500$=1500$ km 

Time taken by usual speed =1500x$={\displaystyle \frac{1500}{x}}$  hr 

Time taken by new speed =1500x+250$={\displaystyle \frac{1500}{x+250}}$ hr 

According to question,

1500x−1500x+250=12${\displaystyle \frac{1500}{x}}-{\displaystyle \frac{1500}{x+250}}={\displaystyle \frac{1}{2}}$ 

⇒1500x+1500×250−1500xx2+250x=12$\Rightarrow {\displaystyle \frac{1500x+1500\times 250-1500x}{x^2+250x}}={\displaystyle \frac{1}{2}}$ 

⇒x2+250x=1500×2502$\Rightarrow x^2+250x={\displaystyle \frac{1500\times 250}{2}}$ 

    ⇒x2+250x=750000$\Rightarrow x^2+250x=750000$ 

    ⇒x2+250x−750000=0$\Rightarrow x^2+250x-750000=0$  

    ⇒x2+1000x−750x−750000=0$\Rightarrow x^2+1000x-750x-750000=0$

⇒x(x+1000)−750(x+1000)=0$\Rightarrow x\left(x+1000\right)-750\left(x+1000\right)=0$ 

⇒x=750orx=−1000$\Rightarrow x=750orx=-1000$ 

Therefore, usual speed is 750$750$ km/hr, −1000$-1000$is neglected as speed cannot be negative.

8. A motor boat, whose speed is 15$15$ km/hr in still water, goes 30$30$ km downstream and comes back in a total time of 4$4$  hr 30$30$ minutes, find the speed of the stream.

Ans: 

Speed of motor boat in still water = 15$=15$ km/hr 

Speed of stream =x$=x$ km/hr 

Speed in downward direction 15+x$15+x$ 

Speed in downward direction 15−x$15-x$ 

 According to question,

3015+x+3015−x=412${\displaystyle \frac{30}{15+x}}+{\displaystyle \frac{30}{15-x}}=4{\displaystyle \frac{1}{2}}$ 

⇒30(15−x)+30(15+x)(15+x)(15−x)=92$\Rightarrow {\displaystyle \frac{30\left(15-x\right)+30\left(15+x\right)}{\left(15+x\right)\left(15-x\right)}}={\displaystyle \frac{9}{2}}$ 

⇒450−30x+450+30x225−x2=92$\Rightarrow {\displaystyle \frac{450-30x+450+30x}{225-x^2}}={\displaystyle \frac{9}{2}}$ 

⇒9(225−x2)=1800$\Rightarrow 9\left(225-x^2\right)=1800$ 

⇒225−x2=200$\Rightarrow 225-x^2=200$ 

∴x=5$\therefore x=5$ 

Speed of stream = 5$=5$ km/hr.

9. A swimming pool is filled with three pipes with uniform flow. The first two pipes operating simultaneously fill the pool in the same time during which the pool is the same time during which the pool is filled by the third pipe alone. The second pipe fills the pool five hours faster than the first pipe and four hours slower than the third pipe. Find the time required by each pipe to fill the pool separately.

Ans:  

Let x$x$ be the number of hours required by the second pipe alone to till the pool and first pipe (x+5)$\left(x+5\right)$ hour while third pipe(x−4)$\left(x-4\right)$ hour

1x+5+1x=1x−4${\displaystyle \frac{1}{x+5}}+{\displaystyle \frac{1}{x}}={\displaystyle \frac{1}{x-4}}$ 

⇒x+x+5x2+5x=1x−4$\Rightarrow {\displaystyle \frac{x+x+5}{x^2+5x}}={\displaystyle \frac{1}{x-4}}$ 

⇒x2−8x−20=0$\Rightarrow x^2-8x-20=0$ 

⇒x2−10x+2x−20=0$\Rightarrow x^2-10x+2x-20=0$ 

⇒x(x−10)+2(x−10)=0$\Rightarrow x\left(x-10\right)+2\left(x-10\right)=0$ 

⇒(x−10)(x+2)=0$\Rightarrow \left(x-10\right)\left(x+2\right)=0$ 

⇒x=10orx=−2(Neglected)$\Rightarrow x=10orx=-2\left(Neglected\right)$ 

10. A two-digit number is such that the product of its digits is18$18$. When 63$63$  is subtracted from the number the digit interchanges their places. Find the number

Ans: 

Let digit on unit’s place =x$=x$ 

Digit on ten’s place =y$=y$ 

 xy=18$xy=18$ (given) 

Number = 10.y+x$y+x$ 

=10(18x)+x$=10\left({\displaystyle \frac{18}{x}}\right)+x$ 

According to question,

  10(18x)+x−63=10x+18x$10\left({\displaystyle \frac{18}{x}}\right)+x-63=10x+{\displaystyle \frac{18}{x}}$ 

   ⇒180x+x−631=10x2+18x$\Rightarrow {\displaystyle \frac{180}{x}}+{\displaystyle \frac{x-63}{1}}={\displaystyle \frac{10x^2+18}{x}}$ 

   ⇒180+x2−63xx=10x2+18x$\Rightarrow {\displaystyle \frac{180+x^2-63x}{x}}={\displaystyle \frac{10x^2+18}{x}}$ 

   ⇒9x2+63x−162=0$\Rightarrow 9x^2+63x-162=0$ 

   ⇒9(x2+9x−2x−18)=0$\Rightarrow 9\left(x^2+9x-2x-18\right)=0$ 

   ⇒x(x+9)−2(x+9)=0$\Rightarrow x\left(x+9\right)-2\left(x+9\right)=0$ 

    ⇒x=2orx=−9$\Rightarrow x=2orx=-9$ 

Number=10(182)+2=92$=10\left({\displaystyle \frac{18}{2}}\right)+2=92$ 

11. A factory kept increasing its output by the same percent ago every year. Find the percentage if it is known that the output is doubled in the last two years.

Ans: 

According to question, 

2P=p(1+r100)2$=p{\left(1+{\displaystyle \frac{r}{100}}\right)}^2$ 

⇒2–√1=1+r100$\Rightarrow {\displaystyle \frac{\sqrt{2}}{1}}=1+{\displaystyle \frac{r}{100}}$ 

2–√−1=r100$\sqrt{2}-1={\displaystyle \frac{r}{100}}$ 

⇒r=(2–√−1)100$\Rightarrow r=\left(\sqrt{2}-1\right)100$ 

12. Two pipes running together can fill a cistern in if one pipe takes 3113$3{\displaystyle \frac{1}{13}}$  minutes more than the other to fill it, find the time in which each pipe would fill the cistern.

Ans: 

Let the faster pipe takes minutes to fill the cistern and the slower pipe will take (x+3)$\left(x+3\right)$  minutes. 

According to question,

1x+1x+3=14013${\displaystyle \frac{1}{x}}+{\displaystyle \frac{1}{x+3}}={\displaystyle \frac{1}{{\displaystyle \frac{40}{13}}}}$ 

⇒1x+1x+3=1340$\Rightarrow {\displaystyle \frac{1}{x}}+{\displaystyle \frac{1}{x+3}}={\displaystyle \frac{13}{40}}$ 

⇒x+3+xx2+3x=1340$\Rightarrow {\displaystyle \frac{x+3+x}{x^2+3x}}={\displaystyle \frac{13}{40}}$ 

⇒13x2−41x−120=0$\Rightarrow 13x^2-41x-120=0$ 

⇒13x2−65x+24x−120=0$\Rightarrow 13x^2-65x+24x-120=0$ 

⇒13x(x−5)+24(x−5)=0$\Rightarrow 13x\left(x-5\right)+24\left(x-5\right)=0$ 

∴x=5orx=−2413(Neglected)$\therefore x=5orx={\displaystyle \frac{-24}{13}}\left(Neglected\right)$ 

13. If the roots of the equation(a−b)x2+(b−c)x+(c−a)=0$\left(a-b\right)x^2+\left(b-c\right)x+\left(c-a\right)=0$  are equal, prove that2a=b+c$2a=b+c$ 

Ans: 

(a−b)x2+(b−c)x+(c−a)=0$\left(a-b\right)x^2+\left(b-c\right)x+\left(c-a\right)=0$

D=b2−4ac$D=b^2-4ac$ 

=(b−c)2−4×(a−b)×(c−a)$={\left(b-c\right)}^2-4\times \left(a-b\right)\times \left(c-a\right)$ 

=b2+c2−2bc−4(ac−a2−bc+ab)$=b^2+c^2-2bc-4\left(ac-a^2-bc+ab\right)$ 

=b2+c2−2bc−4ac+4a2+4bc−4ab$=b^2+c^2-2bc-4ac+4a^2+4bc-4ab$ 

=(b)2+(c)2+(2a)2+2bc−4ac−4ab$={\left(b\right)}^2+{\left(c\right)}^2+{\left(2a\right)}^2+2bc-4ac-4ab$ 

=(b+c−2a)2$={\left(b+c-2a\right)}^2$ 

For equal root s,

 D = 0

⇒(b+c−2a)2=0$\Rightarrow {\left(b+c-2a\right)}^2=0$ 

⇒(b+c−2a)=0$\Rightarrow \left(b+c-2a\right)=0$ 

⇒b+c=2a$\Rightarrow b+c=2a$ 

14. Two circles touch internally. The sum of their areas is 116πcm2$116\pi c{m}^2$ and the distance between their centers is 6$6$ cm. Find the radii of the circles.

Ans: 

Let r1$r_1$ and r2$r_2$ be the radius of two circles 

According to question,

    Πr12+Πr22=116Π$\mathrm{\Pi }{r_1}^2+\mathrm{\Pi }{r_2}^2=116\mathrm{\Pi }$ 

    ⇒r12+r22=116......(i)$\Rightarrow {r_1}^2+{r_2}^2=\mathrm{116......}\left(i\right)$ 

    r2−r1=6$r_2-r_1=6$ (Given)

(Image will be uploaded soon)

   ⇒r2=6+r1$\Rightarrow r_2=6+r_1$ 

   Puting the value ofr2$r_2$  in eq. … (i) we get

   ⇒r12+36+r12+12r1=116$\Rightarrow {r_1}^2+36+{r_1}^2+12r_1=116$ 

   ⇒2r12+12r1−80=0$\Rightarrow 2{r_1}^2+12r_1-80=0$ 

  r12+6r1−40=0${r_1}^2+6r_1-40=0$ 

  ⇒r12+10r1−4r1−40=0$\Rightarrow {r_1}^2+10r_1-4r_1-40=0$ 

  ⇒r1(r1+10)−4(r1+10)=0$\Rightarrow r_1\left(r_1+10\right)-4\left(r_1+10\right)=0$ 

  ⇒(r1+10)(r1−4)=0$\Rightarrow \left(r_1+10\right)\left(r_1-4\right)=0$ 

  whenr1=4cm$when{r}_1=4cm$ 

  r2=6+r1$r_2=6+r_1$ 

  =6+4$=6+4$ 

  r2=10cm$r_2=10cm$ 

  ⇒r1=−10$\Rightarrow r_1=-10$ (Neglect) or r1= 4 cm$r_1=4cm$ 

15. A piece of cloth costs Rs.200$200$. If the piece was 5$5$ m longer and each metre of cloth costs Rs. 2$2$  less the cost of the piece would have remained unchanged. How long is the  piece and what is the original rate per meter?

Ans: 

Let the length of piece =x$=x$  m 

Rate per meter =200x$={\displaystyle \frac{200}{x}}$  

A New length = (x+5)$=\left(x+5\right)$ 

A New rate per meter =200x+5$={\displaystyle \frac{200}{x+5}}$ 

According to question,

200x−200x+5=2${\displaystyle \frac{200}{x}}-{\displaystyle \frac{200}{x+5}}=2$ 

⇒200(x+5)−200x(x+5)=21$\Rightarrow {\displaystyle \frac{200\left(x+5\right)-200x}{\left(x+5\right)}}={\displaystyle \frac{2}{1}}$ 

⇒200x+1000−200xx2+5x=21$\Rightarrow {\displaystyle \frac{200x+1000-200x}{x^2+5x}}={\displaystyle \frac{2}{1}}$ 

⇒x2+5x=500$\Rightarrow x^2+5x=500$ 

⇒x2+25x−20x−500=0$\Rightarrow x^2+25x-20x-500=0$ 

⇒x(x+25)−20(x+25)=0$\Rightarrow x\left(x+25\right)-20\left(x+25\right)=0$ 

⇒(x+25)(x−20)=0$\Rightarrow \left(x+25\right)\left(x-20\right)=0$ 

∴x=−25(Neglect)orx=20$\therefore x=-25\left(Neglect\right)orx=20$ 

Rate per meter = 10$=10$ 

16. ax2+bx+x = 0$a{x}^2+bx+x=0$ , a≠0$a\ne 0$ solve by quadratic formula.

Ans:

ax2+bx+x = 0$a{x}^2+bx+x=0$ 

⇒x2+bax+ca=0$\Rightarrow x^2+{\displaystyle \frac{b}{a}}x+{\displaystyle \frac{c}{a}}=0$ 

⇒x2+bax(b2a)2−(b2a)2+ca=0$\Rightarrow x^2+{\displaystyle \frac{b}{a}}x{\left({\displaystyle \frac{b}{2a}}\right)}^2-{\left({\displaystyle \frac{b}{2a}}\right)}^2+{\displaystyle \frac{c}{a}}=0$ 

⇒(x+b2a)2=b24a2−ca$\Rightarrow {\left(x+{\displaystyle \frac{b}{2a}}\right)}^2={\displaystyle \frac{b^2}{4a^2}}-{\displaystyle \frac{c}{a}}$ 

x⇒(x+b2a)2=b2−4ac4a2$\Rightarrow {\left(x+{\displaystyle \frac{b}{2a}}\right)}^2={\displaystyle \frac{b^2-4ac}{4a^2}}$ 

⇒x+b2a=±b2−4ac4a2−−−−−−−√$\Rightarrow x+{\displaystyle \frac{b}{2a}}=\pm \sqrt{{\displaystyle \frac{b^2-4ac}{4a^2}}}$ 

⇒x+−b2a=±b2−4ac2a−−−−−−−√$\Rightarrow x+{\displaystyle \frac{-b}{2a}}=\pm \sqrt{{\displaystyle \frac{b^2-4ac}{2a}}}$ 

⇒x=−b±b2−4ac−−−−−−−√2a$\Rightarrow x={\displaystyle \frac{-b\pm \sqrt{b^2-4ac}}{2a}}$

⇒x=−b+b2−4ac−−−−−−−√2a$\Rightarrow x={\displaystyle \frac{-b+\sqrt{b^2-4ac}}{2a}}$ 

or⇒x=−b−b2−4ac−−−−−−−√2a$or\Rightarrow x={\displaystyle \frac{-b-\sqrt{b^2-4ac}}{2a}}$ 

17. The length of the hypotenuse of a right-angled triangle exceeds the length of the base by 2$2$ cm and exceeds twice the length of the altitude by 1$1$  cm. Find the length of each side of the triangle.

Ans: 

Let base of the triangle = x$=x$ 

Altitude of the triangle = y$=y$ 

Hypotenuse of the triangle = h$=h$ 

(Image will be uploaded soon)

According to question,

h=x+2$h=x+2$ 

h=2y+1$h=2y+1$ 

⇒x+2=2y+1$\Rightarrow x+2=2y+1$ 

    ⇒x+2−1=2y$\Rightarrow x+2-1=2y$ 

⇒x−1=2y$\Rightarrow x-1=2y$ 

⇒x−12=y$\Rightarrow {\displaystyle \frac{x-1}{2}}=y$ 

Andx2+y2=h2$And{x}^2+y^2=h^2$ 

⇒x2+(x−12)2=(x+2)2$\Rightarrow x^2+{\left({\displaystyle \frac{x-1}{2}}\right)}^2={\left(x+2\right)}^2$ 

⇒x2−15x+x−15=0$\Rightarrow x^2-15x+x-15=0$ 

⇒(x−15)(x+1)=0$\Rightarrow \left(x-15\right)\left(x+1\right)=0$ 

⇒x=15orx=−1$\Rightarrow x=15orx=-1$ 

Base of the triangle = 15$=15$ cm

 Altitude of the triangle =x+12=8cm$={\displaystyle \frac{x+1}{2}}=8cm$ 

Hypotenuse of the triangle = 17$=17$ cm

18. Find the roots of the following quadratic equations if they exist by the method of completing square.

i). 2x2 −7x+3=0$2x^2-7x+3=0$ 

Ans: 

(i) 2x2 −7x+3=0$2x^2-7x+3=0$

Dividing the equation by 2$2$  to make coefficient of x2$x^2$ equal to1$1$ we get

x2−72x+32=0$x^2-{\displaystyle \frac{7}{2}}x+{\displaystyle \frac{3}{2}}=0$ 

Dividing the middle term of the equation by2x$2x$, we get

72x×12x=74${\displaystyle \frac{7}{2}}x\times {\displaystyle \frac{1}{2x}}={\displaystyle \frac{7}{4}}$ 

Adding and subtracting square of 74${\displaystyle \frac{7}{4}}$ from the equation x2−72x+32=0$x^2-{\displaystyle \frac{7}{2}}x+{\displaystyle \frac{3}{2}}=0$ we get

x2−72x+32+(74)2−(74)2=0$x^2-{\displaystyle \frac{7}{2}}x+{\displaystyle \frac{3}{2}}+{\left({\displaystyle \frac{7}{4}}\right)}^2-{\left({\displaystyle \frac{7}{4}}\right)}^2=0$

⇒x2+(74)2−72x+32+−(74)2=0$\Rightarrow x^2+{\left({\displaystyle \frac{7}{4}}\right)}^2-{\displaystyle \frac{7}{2}}x+{\displaystyle \frac{3}{2}}+-{\left({\displaystyle \frac{7}{4}}\right)}^2=0$      {(a−b)2=a2+b2−2ab}$\left\{{\left(a-b\right)}^2=a^2+b^2-2ab\right\}$ 

⇒(x-74)2+24-4916=0$\Rightarrow {\left(\text{x-}{\displaystyle \frac{\text{7}}{\text{4}}}\right)}^{\text{2}}\text{+}{\displaystyle \frac{\text{24-49}}{\text{16}}}\text{=0}$ 

⇒(x-74)2=49-2416$\Rightarrow {\left(\text{x-}{\displaystyle \frac{\text{7}}{\text{4}}}\right)}^{\text{2}}\text{=}{\displaystyle \frac{\text{49-24}}{\text{16}}}$ 

Square rooting on both the sides we get

⇒x-74= ± 54$\Rightarrow \text{x-}{\displaystyle \frac{\text{7}}{\text{4}}}\text{= }\pm {\displaystyle \frac{\text{5}}{\text{4}}}$ 

⇒x=54+74=124=3andx=-54+74=24=12$\Rightarrow x\text{=}{\displaystyle \frac{\text{5}}{\text{4}}}\text{+}{\displaystyle \frac{\text{7}}{\text{4}}}\text{=}{\displaystyle \frac{\text{12}}{\text{4}}}\text{=3}\text{and}\text{x=-}{\displaystyle \frac{\text{5}}{\text{4}}}\text{+}{\displaystyle \frac{\text{7}}{\text{4}}}\text{=}{\displaystyle \frac{\text{2}}{\text{4}}}\text{=}{\displaystyle \frac{\text{1}}{\text{2}}}$ 

Therefore,x=12,3$x={\displaystyle \frac{1}{2}},3$ 

ii). 2x2+x 4=0$2x^2+x4=0$ 

Ans:

2x2+x-- 4=0$\text{2}{\text{x}}^{\text{2}}\text{+x-- 4=0}$

Dividing equation by2$2$ , 

x2+x2-2=0${\text{x}}^{\text{2}}\text{+}{\displaystyle \frac{\text{x}}{\text{2}}}\text{-2=0}$ 

Following procedure of completing square,

x2+x2-2+(14)2-(14)2=0${\text{x}}^{\text{2}}\text{+}{\displaystyle \frac{\text{x}}{\text{2}}}\text{-2+}{\left({\displaystyle \frac{\text{1}}{\text{4}}}\right)}^{\text{2}}\text{-}{\left({\displaystyle \frac{\text{1}}{\text{4}}}\right)}^{\text{2}}\text{=0}$ 

⇒x2+x2+(14)2-2-116=0$\Rightarrow {\text{x}}^{\text{2}}\text{+}{\displaystyle \frac{\text{x}}{\text{2}}}\text{+}{\left({\displaystyle \frac{\text{1}}{\text{4}}}\right)}^{\text{2}}\text{-2-}{\displaystyle \frac{\text{1}}{\text{16}}}\text{=0}$      {(a-b)2=a2+b2-2ab}$\left\{{\left(\text{a-b}\right)}^{\text{2}}\text{=}{\text{a}}^{\text{2}}\text{+}{\text{b}}^{\text{2}}\text{-2ab}\right\}$

⇒(x+14)2-3316=0$\Rightarrow {\left(\text{x+}{\displaystyle \frac{\text{1}}{\text{4}}}\right)}^{\text{2}}\text{-}{\displaystyle \frac{\text{33}}{\text{16}}}\text{=0}$ 

⇒(x+14)2-3316$\Rightarrow {\left(\text{x+}{\displaystyle \frac{\text{1}}{\text{4}}}\right)}^{\text{2}}\text{-}{\displaystyle \frac{\text{33}}{\text{16}}}$

Taking square root on both sides,

⇒x+14= ± 33−−√4$\Rightarrow \text{x+}{\displaystyle \frac{\text{1}}{\text{4}}}\text{= }\pm {\displaystyle \frac{\sqrt{\text{33}}}{\text{4}}}$ 

⇒x=33−−√4−14=33−−√-14andx=-33−−√4-14=-33−−√-14$\Rightarrow x={\displaystyle \frac{\sqrt{33}}{4}}-{\displaystyle \frac{1}{4}}\text{=}{\displaystyle \frac{\sqrt{\text{33}}\text{-1}}{\text{4}}}\text{and}\text{x=-}{\displaystyle \frac{\sqrt{\text{33}}}{\text{4}}}\text{-}{\displaystyle \frac{\text{1}}{\text{4}}}\text{=}{\displaystyle \frac{\text{-}\sqrt{\text{33}}\text{-1}}{\text{4}}}$ 

Therefore,x=33−−√−14,−33−−√−14$\text{x}={\displaystyle \frac{\sqrt{33}-1}{4}},{\displaystyle \frac{-\sqrt{33}-1}{4}}$ 

iii). 4x2+43–√x+3=0$4x^2+4\sqrt{3}x+3=0$ 

Ans:

4x2+43–√x+3=0$\text{4}{\text{x}}^{\text{2}}\text{+4}\sqrt{\text{3}}\text{x+3=0}$

Dividing this equation by 4,$4,$ we get

x2+3–√x+34=0${\text{x}}^{\text{2}}\text{+}\sqrt{\text{3}}\text{x+}{\displaystyle \frac{\text{3}}{\text{4}}}\text{=0}$ 

By the procedure of completing square we get

⇒x2+3–√x+34+(3–√2)2-(3–√2)2=0$\Rightarrow {\text{x}}^{\text{2}}\text{+}\sqrt{\text{3}}\text{x+}{\displaystyle \frac{\text{3}}{\text{4}}}\text{+}{\left({\displaystyle \frac{\sqrt{\text{3}}}{\text{2}}}\right)}^{\text{2}}\text{-}{\left({\displaystyle \frac{\sqrt{\text{3}}}{\text{2}}}\right)}^{\text{2}}\text{=0}$ 

⇒x2+(3–√2)2+3–√x+34-34=0$\Rightarrow {\text{x}}^{\text{2}}\text{+}{\left({\displaystyle \frac{\sqrt{\text{3}}}{\text{2}}}\right)}^{\text{2}}\text{+}\sqrt{\text{3}}\text{x+}{\displaystyle \frac{\text{3}}{\text{4}}}\text{-}{\displaystyle \frac{\text{3}}{\text{4}}}\text{=0}$       {(a-b)2=a2+b2-2ab}$\left\{{\left(\text{a-b}\right)}^{\text{2}}\text{=}{\text{a}}^{\text{2}}\text{+}{\text{b}}^{\text{2}}\text{-2ab}\right\}$

⇒(x+3–√2)2=0$\Rightarrow {\left(\text{x+}{\displaystyle \frac{\sqrt{\text{3}}}{\text{2}}}\right)}^{\text{2}}\text{=0}$ 

⇒(x+3–√2)(x+3–√2)=0$\Rightarrow \left(\text{x+}{\displaystyle \frac{\sqrt{\text{3}}}{\text{2}}}\right)\left(\text{x+}{\displaystyle \frac{\sqrt{\text{3}}}{\text{2}}}\right)\text{=0}$ 

⇒x+3–√2=0,x+3–√2=0$\Rightarrow \text{x+}{\displaystyle \frac{\sqrt{\text{3}}}{\text{2}}}\text{=0,x+}{\displaystyle \frac{\sqrt{\text{3}}}{\text{2}}}\text{=0}$ 

⇒x=-3–√2,-3–√2$\Rightarrow \text{x=-}{\displaystyle \frac{\sqrt{\text{3}}}{\text{2}}}\text{,-}{\displaystyle \frac{\sqrt{\text{3}}}{\text{2}}}$

iv). 2x2+x+4=0$2x^2+x+4=0$

Ans:

2x2+x+4=0$\text{2}{\text{x}}^{\text{2}}\text{+x+4=0}$

Dividing this equation by2$2$we get

x2+x2+2=0${\text{x}}^{\text{2}}\text{+}{\displaystyle \frac{\text{x}}{\text{2}}}\text{+2=0}$ 

By the procedure of completing square,

⇒x2+x2+2+(14)2−(14)2=0$\Rightarrow x^2+{\displaystyle \frac{x}{2}}+2+{\left({\displaystyle \frac{1}{4}}\right)}^2-{\left({\displaystyle \frac{1}{4}}\right)}^2=0$ 

⇒x2+(14)2+x2+2-(14)2=0$\Rightarrow {\text{x}}^{\text{2}}\text{+}{\left({\displaystyle \frac{\text{1}}{\text{4}}}\right)}^{\text{2}}\text{+}{\displaystyle \frac{\text{x}}{\text{2}}}\text{+2-}{\left({\displaystyle \frac{\text{1}}{\text{4}}}\right)}^{\text{2}}\text{=0}$      {(a−b)2=a2+b2−2ab}$\left\{{\left(a-b\right)}^2=a^2+b^2-2ab\right\}$

⇒(x+14)2+2−116=0$\Rightarrow {\left(x+{\displaystyle \frac{1}{4}}\right)}^2+2-{\displaystyle \frac{1}{16}}=0$ 

(x+14)2=116−2=1−3216${\left(x+{\displaystyle \frac{1}{4}}\right)}^2={\displaystyle \frac{1}{16}}-2={\displaystyle \frac{1-32}{16}}$ 

Right hand side does not exist because the square root of the negative number does not exist. 

Therefore, there is no solution for quadratic equation 2x2+x+4=0$2x^2+x+4=0$