ML Aggarwal Solution Class 10 Chapter 5 Quadratic Equations in One Variable Exercise 5.3

 Exercise 5.3

Question 1

(i) 2x² – 7x + 6 = 0

(ii) 2x² – 6x + 3 = 0

Sol :

(i) 2x² – 7x + 6 = 0

Here a = 2, b = -7, c = 6

∴$\mathrm{D}=b^{2}-4 a c=(-7)^{2}-4 \times 2 \times 6$

=49-48=1

∴$x=\frac{-b \pm \sqrt{b^{2}-4 a c}}{2 a}=\frac{-b \pm \sqrt{D}}{2 a}$

$=\frac{-(-7) \pm \sqrt{1}}{2 \times 2}=\frac{7 \pm 1}{4}$

∴$x_{1}=\frac{7+1}{4}=\frac{8}{4}=2$ and $x_{2}=\frac{7-1}{4}=\frac{6}{4}=\frac{3}{2}$

∴$x=2 ,\frac{3}{2}$

(ii) $2 x^{2}-6 x+3=0$

Here a=2, b=-6, c=3

then $D=b^{2}-4 a c=(-6)^{2}-4 \times 2 \times 3$

=36-24=12

Now $x=\frac{-b \pm \sqrt{D}}{2 a}=\frac{-(-6) \pm \sqrt{12}}{2 \times 2}=\frac{6 \pm 2 \sqrt{3}}{4}$

∴$x_{1}=\frac{6+2 \sqrt{3}}{4}=\frac{2(3+\sqrt{3})}{4}=\frac{3+\sqrt{3}}{2}$

$x_{2}=\frac{6-2 \sqrt{3}}{4}=\frac{2(3-\sqrt{3})}{4}=\frac{3-\sqrt{3}}{2}$

Hence $x=\frac{3+\sqrt{3}}{2}, \frac{3-\sqrt{3}}{2}$

Question 2

(i) x² + 7x – 7 = 0

(ii) (2x + 3)(3x – 2) + 2 = 0

Sol :

(i) x² + 7x – 7 = 0

Here a = 1, b = 7, c = -7

$\therefore D=b^{2}-4 a c=(7)^{2}-4 \times 1(-7)$

=49+28=77

$\because x=\frac{-b \pm \sqrt{D}}{2 a}=\frac{-7 \pm \sqrt{77}}{2 \times 1}=\frac{-7 \pm \sqrt{77}}{2}$

$\therefore x_{1}=\frac{-7+\sqrt{77}}{2}$ and $x_{2}=\frac{-7-\sqrt{77}}{2}$

Hence $x=\frac{-7+\sqrt{77}}{2}, \frac{-7-\sqrt{77}}{2}$

(ii)

(2 x+3)(3 x-2)+2=0

$6x^{2}-4 x+9x-6+2=0$

$6 x^{2}+5 x-4=0$

Here a=6, b=5, c=-4

$\mathrm{D}=b^{2}-4 a c=(5)^{2}-4 \times 6 \times(-4)$

=25+96=121

$\because x=\frac{-b \pm \sqrt{\mathrm{D}}}{2 a}=\frac{-5 \pm \sqrt{121}}{2 \times 6}=\frac{-5 \pm 11}{12}$

$\therefore x_{1}=\frac{-5+11}{12}=\frac{6}{12}=\frac{1}{2}$

$x_{2}=\frac{-5-11}{12}=\frac{-16}{12}=\frac{-4}{3}$

Hence $x=\frac{1}{2}, \frac{-4}{3}$ Ans.

Question 3

(i)256x² – 32x + 1 = 0

(ii) 25x² + 30x + 7 = 0

Sol :

(i) 256x² – 32x + 1 = 0

Here a = 256, b = -32, c = 1

$\mathrm{D}=b^{2}-4 a c=(-32)^{2}-4 \times 256 \times 1$

=1024-1024=0

$\because x=\frac{-b \pm \sqrt{D}}{2 a}=\frac{-(-32) \pm \sqrt{0}}{2 \times 256}=\frac{32}{512}=\frac{1}{16}$

$x_{1}=\frac{1}{16}, x_{2}=\frac{1}{16}$

Hence $x=\frac{1}{16}, \frac{1}{16}$ Ans.

(ii) $25 x^{2}+30 x+7=0$

Here $a=25, b=30, c=7$

$D=b^{2}-4 a c=(30)^{2}-4 \times 25 \times 7$

=900-700=200

$\because x=\frac{-b \pm \sqrt{D}}{2 a}=\frac{-30 \pm \sqrt{200}}{2 \times 25}$

$=\frac{-30 \pm \sqrt{100 \times 2}}{50}=\frac{-30 \pm 10 \sqrt{2}}{50}=\frac{-3 \pm \sqrt{2}}{5}$

$\therefore x_{1}=\frac{-3+\sqrt{2}}{5}$ and $x_{2}=\frac{-3-\sqrt{2}}{5}$

Hence $x=\frac{-3+\sqrt{2}}{5}, \frac{-3-\sqrt{2}}{5}$

Question 4

(i) 2x² + √5x – 5 = 0

(ii) √3x² + 10x – 8√3 = 0

Sol :

(i) 2x² + √5x – 5 = 0

Here a = 2, b = √5, c = -5

$D=b^{2}-4 a c=(\sqrt{5})^{2}-4 \times 2 \times(-5)$

=5+40=45

$\because x=\frac{-b \pm \sqrt{\mathrm{D}}}{2 a}=\frac{-\sqrt{5} \pm \sqrt{45}}{2 \times 2}$

$=\frac{-\sqrt{5} \pm \sqrt{9 \times 5}}{4}=\frac{-\sqrt{5} \pm 3 \sqrt{5}}{4}$

$\therefore x_{1}=\frac{-\sqrt{5}+3 \sqrt{5}}{4}=\frac{2 \sqrt{5}}{4}=\frac{\sqrt{5}}{2}$

$x_{2}=\frac{-\sqrt{5}-3 \sqrt{5}}{4}=\frac{-4 \sqrt{5}}{4}=-\sqrt{5}$

Hence $x=\frac{\sqrt{5}}{2},-\sqrt{5}$ Ans.

(ii) $\sqrt{3} x^{2}+10 x-8 \sqrt{3}=0$

Here $a=\sqrt{3}, b=10, c=-8 \sqrt{3}$

$D=b^{2}-4 a c=(10)^{2}-4 \times \sqrt{3} \times(-8 \sqrt{3})$

=100+96=196

$\because x=\frac{-b \pm \sqrt{D}}{2 a}=\frac{-10 \pm \sqrt{196}}{2 \times \sqrt{3}}=\frac{-10 \pm 14}{2 \sqrt{3}}$

$\therefore x_{1}=\frac{-10+14}{2 \sqrt{3}}=\frac{4}{2 \sqrt{3}}=\frac{2 \times \sqrt{3}}{\sqrt{3} \times \sqrt{3}}=\frac{2 \sqrt{3}}{3}$

$x_{2}=\frac{-10-14}{2 \sqrt{3}}=\frac{-24}{2 \sqrt{3}}=\frac{-12 \times \sqrt{3}}{\sqrt{3} \times \sqrt{3}}$

$=\frac{-12 \sqrt{3}}{3}=-4 \sqrt{3}$

Hence $x=\frac{2 \sqrt{3}}{3},-4 \sqrt{3}$ Ans.

Question 5

(i) $\frac{x-2}{x+2}+\frac{x+2}{x-2}=4$

(ii) $\frac{x+1}{x+3}=\frac{3 x+2}{2 x+3}$

Sol :

(i) $\frac{x-2}{x+2}+\frac{x+2}{x-2}=4$

$\Rightarrow \frac{(x-2)^{2}+(x+2)^{2}}{(x+2)(x-2)}=4$

$\Rightarrow \frac{x^{2}-4 x+4+x^{2}+4 x+4}{x^{2}-4}=4$

$\Rightarrow 2 x^{2}+8=4 x^{2}-16$

$\Rightarrow 2 x^{2}+8-4 x^{2}+16=0$

$\Rightarrow-2 x^{2}+24=0$

$\Rightarrow \quad x^{2}-12=0$

Here a=1, b=0, c=-12

$\mathrm{D}=b^{2}-4 a c=(0)^{2}-4 \times 1(-12)$

=0+48=48

$\because x=\frac{-b \pm \sqrt{D}}{2 a}=\frac{0 \pm \sqrt{48}}{2 \times 1}=\frac{\pm \sqrt{48}}{2}$

$=\frac{\pm \sqrt{16 \times 3}}{2}=\pm \frac{4 \sqrt{3}}{2}=\pm 2 \sqrt{3}$

Hence roots are $-2 \sqrt{3},-2 \sqrt{3}$ Ans.

(ii) $\frac{x+1}{x+3}=\frac{3 x+2}{2 x+3}$

$(x+1)(2 x+3)=(3 x+2)(x+3)$

$\Rightarrow 2 x^{2}+3 x+2 x+3$

$=3 x^{2}+9 x+2 x+6$

$\Rightarrow \quad 2 x^{2}+5 x+3-3 x^{2}-11 x-6=0$

$\Rightarrow \quad-x^{2}-6 x-3=0$

$\Rightarrow x^{2}+6 x+3=0$

Here a=1, b=6, c=3

$\mathrm{D}=b^{2}-4 a c=(6)^{2}-4 \times 1 \times 3$

$=36-12=24$

$\because x=\frac{-b \pm \sqrt{\mathrm{D}}}{2 a}$

$=\frac{-6 \pm \sqrt{24}}{2 \times 1}$

$\therefore x_{1}=-3+\sqrt{6}, x_{2}=-3-\sqrt{6}$

Hence $x=-3+\sqrt{6},-3-\sqrt{6}$ Ans.

Question 6

(i) a (x² + 1) = (a² + 1) x , a ≠ 0

(ii) 4x² – 4ax + (a² – b²) = 0

Sol :

(i) a (x² + 1) = (a² + 1) x

ax² – (a² + 1)x + a = 0

Here $a=a, b=-\left(a^{2}+1\right), c=a$

$\mathrm{D}=b^{2}-4 a c=\left[-\left(a^{2}+1\right)\right]^{2}-4 \times a \times a$

$=a^{4}+2 a^{2}+1-4 a^{2}=a^{4}-2 a^{2}+1$

$=\left(a^{2}-1\right)^{2}$

$\because x=\frac{-b \pm \sqrt{\mathrm{D}}}{2 a}$

$=\frac{\left(a^{2}+1\right) \pm \sqrt{\left(a^{2}-1\right)^{2}}}{2 \times a}=\frac{\left(a^{2}+1\right) \pm\left(a^{2}-1\right)}{2 a}$

$\therefore x_{1}=\frac{a^{2}+1+a^{2}-1}{2 a}=\frac{2 a^{2}}{2 a}=a$

$x_{2}=\frac{a^{2}+1-a^{2}+1}{2 a}=\frac{2}{2 a}=\frac{1}{a}$

Hence $x=a, \frac{1}{a}$. Ans.

(ii) $4 x^{2}-4 a x+\left(a^{2}-b^{2}\right)=0$

Here $a=4, b=-4 a, c=a^{2}-b^{2}$

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

$=16 a^{2}-16\left(a^{2}-b^{2}\right)$

$=16 a^{2}-16 a^{2}+16 b^{2}$

$D=16 b^{2}$

$\therefore x=\frac{-b \pm \sqrt{D}}{2 a}=\frac{-(-4 a) \pm \sqrt{16 b^{2}}}{2 \times 4}$

$=\frac{4 a \pm 4 b}{8}=\frac{a \pm b}{2}$

$\therefore x_{1}=\frac{a+b}{2}, x_{2}=\frac{a-b}{2}$

Hence $x=\frac{a+b}{2}, \frac{a-b}{2}$ Ans.

Question 7

(i) $x-\frac{1}{x}=3, x \neq 0$

(ii) $\frac{1}{x}+\frac{1}{x-2}=3, x \neq 0,2$

Sol :

(i) $x-\frac{1}{x}=3$

$x^{2}-1=3 x$

$\Rightarrow x^{2}-3 x-1=0$

Here a=1, b=-3, c=-1

$\therefore b^{2}-4 a c=(-3)^{2}-4 \times 1 \times(-1)$

=9+4=13

$x=\frac{-b \pm \sqrt{b^{2}-4 a c}}{2 a}$

$\therefore  x=\frac{3+\sqrt{13}}{2}$ and $\frac{3-\sqrt{13}}{2}$

(ii) $\frac{1}{x}+\frac{1}{x-2}=3$

$\quad \frac{x-2+x}{x(x-2)}=3 \Rightarrow \frac{2 x-2}{x^{2}-2 x}=3$

$\Rightarrow 3 x^{2}-6 x=2 x-2 \Rightarrow 3 x^{2}-6 x-2 x+2=0$

$\Rightarrow 3 x^{2}-8 x+2=0$

Here a=3, b=-8, c=2

$b^{2}-4 a c=(-8)^{2}-4 \times 3 \times 2$

=64-24=40

$x=\frac{-b \pm \sqrt{b^{2}-4 a c}}{2 a}$

$=\frac{-(-8) \pm \sqrt{40}}{2 \times 3}=\frac{8 \pm 2 \sqrt{10}}{6}=\frac{4 \pm \sqrt{10}}{3}$

$\therefore x=\frac{4+\sqrt{10}}{3}$ and $\frac{4-\sqrt{10}}{3}$

Question 8

$\frac{1}{x-2}+\frac{1}{x-3}+\frac{1}{x-4}=0$

Sol :

$\frac{1}{x-2}+\frac{1}{x-3}+\frac{1}{x-4}=0$

$\Rightarrow \frac{1}{x-2}+\frac{1}{x-3}=-\frac{1}{x-4}$

$\Rightarrow \frac{x-3+x-2}{(x-2)(x-3)}=-\frac{1}{x-4}$

$\Rightarrow \frac{2 x-5}{x^{2}-5 x+6}=\frac{-1}{x-4}$

$\quad(2 x-5)(x-4)=-1\left(x^{2}-5 x+6\right)$

$\Rightarrow 2 x^{2}-8 x-5 x+20=-x^{2}+5 x-6$

$\Rightarrow 2 x^{2}-8 x-5 x+20+x^{2}-5 x+6=0$

$\Rightarrow 3 x^{2}-18 x+26=0$

Here, a=3, b=-18, c=26

$\therefore x=\frac{-b \pm \sqrt{b^{2}-4 a c}}{2 a}$

$=\frac{-(-18) \pm \sqrt{(-18)^{2}-4 \times 3 \times 26}}{2 \times 3}$

$=\frac{18 \pm \sqrt{324-312}}{6}$

$=\frac{18 \pm \sqrt{12}}{6}=\frac{18 \pm 2 \sqrt{3}}{6}$

$=\frac{9 \pm \sqrt{3}}{3}$ (dividing by 2)

$\therefore x=\frac{9+\sqrt{3}}{3}, \frac{9-\sqrt{3}}{3}$

$=3+\frac{\sqrt{3}}{3}, 3-\frac{\sqrt{3}}{3}$

$=3+\frac{1}{\sqrt{3}}, 3-\frac{1}{\sqrt{3}}$

Question 9

Solve : $x: 2\left(\frac{2 x-1}{x+3}\right)-3\left(\frac{x+3}{2 x-1}\right)=5, x \neq-3, \frac{1}{2}$

Sol :

$x: 2\left(\frac{2 x-1}{x+3}\right)-3\left(\frac{x+3}{2 x-1}\right)=5$

Let $\frac{2 x-1}{x+3}=y$ then $\frac{x+3}{2 x-1}=\frac{1}{y}$

$\therefore 2 y-\frac{3}{y}=5$

$2 y^{2}-3=5 y$

$\Rightarrow 2 y^{2}-5 y-3=0$

Here, a=2, b=-5, c=-3

$b^{2}-4 a c=(-5)^{2}-4 \times 2 \times(-3)$

=25+24=49

Now, $y=\frac{-b \pm \sqrt{b^{2}-4 a c}}{2 a}$

$=\frac{-(-5) \pm \sqrt{49}}{2 \times 2}=\frac{5 \pm 7}{4}$

$y=\frac{5+7}{4}=\frac{12}{4}=3$

or $y=\frac{5-7}{4}=\frac{-2}{4}=\frac{-1}{2}$

$\therefore y=3, \frac{-1}{2}$

When $y=3,$ then $\frac{2 x-1}{x+3}=3$

$\Rightarrow 3 x+9=2 x-1$

$\Rightarrow 3 x-2 x=-1-9 \Rightarrow x=-10$

When $y=\frac{-1}{2},$ then

or $\frac{2 x-1}{x+3}=\frac{-1}{2}$

4x-2=-x-3

$4 x+x=-3+2 \Rightarrow 5 x=-1$

$\quad x=\frac{-1}{5}$

$\therefore x=-10, \frac{-1}{5}$

Question 10

Solve the following equation by using quadratic equations for x and give your

(i) x² – 5x – 10 = 0

(ii) 5x(x + 2) = 3

Sol :

(i) x² – 5x – 10 = 0

On comparing with, ax² + bx + c = 0

a=1, b=-5, c=-10

$\because x=\frac{-b \pm \sqrt{b^{2}-4 a c}}{2 a}$

$x=\frac{-(-5) \pm \sqrt{(-5)^{2}-4(1)(-10)}}{2 \times 1}$

$\therefore x=\frac{5 \pm \sqrt{25+40}}{2}$

$\Rightarrow x=\frac{5 \pm \sqrt{65}}{2}=\frac{5 \pm 8.06}{2}$

Either $x=\frac{5+8.06}{2}=\frac{13.06}{2}=6.53$

or $x=\frac{5-8.06}{2}=\frac{-3.06}{2}=-1.53$

∴ x=6.53, x=-1.53

(ii)

5x(x+2)=3

5x(x+2)=3

$5 x^{2}+10 x=3$

$5 x^{2}+10 x-3=0$

Here a=5, b=10, c=-3

$\mathrm{D}=b^{2}-4 a c=(10)^{2}-4 \times 5 \times(-3)$

=100+60=160

$\therefore x=\frac{-b \pm \sqrt{b^{2}-4 a c}}{2 a}=\frac{-10 \pm \sqrt{160}}{2 \times 5}$

$=\frac{-10 \pm \sqrt{16 \times 10}}{10}=\frac{-10 \pm 4 \sqrt{10}}{10}$

$=\frac{-10 \pm 4(3.162)}{10}=\frac{-10 \pm 12.648}{10}$

$\therefore x_{1}=\frac{-10+12.648}{10}=\frac{2.648}{10}=0.2648$

=0.265

$x_{2}=\frac{-10-12.648}{10}=\frac{-22.648}{10}=-2.2648$

$\therefore  x=0.26,-2.26$ Ans.

Question 11

Solve the following equations by using quadratic formula and give your answer correct to 2 decimal places 

(i) 4x² – 5x – 3 = 0

(ii) $2 x-\frac{1}{x}=1$

Sol :

(i) Given equation 4x² – 5x – 3 = 0

Comparing with ax² + bx + c = 0, we have

a=4, b=-5, c=-3

$\therefore x=\frac{-b \pm \sqrt{b^{2}-4 a c}}{2 a}$

$=\frac{-(-5) \pm \sqrt{(-5)^{2}-4 \times 4 \times(-3)}}{2 \times 4}$

$=\frac{5 \pm \sqrt{25+48}}{8}=\frac{5 \pm \sqrt{73}}{8}=\frac{5 \pm 8.544}{8}$

$=\frac{5+8.544}{8}$ or $\frac{5-8.544}{8}$

$=\frac{13.544}{8}$ or $\frac{-3.544}{8}$

$=1.693$ or -0.443

=1.69 or -0.44 (correct to 2 decimal places)

(ii)

$2 x-\frac{1}{x}=7 \Rightarrow 2 x^{2}-1=7 x$

$\Rightarrow 2 x^{2}-7 x-1=0$...(i)

Comparing (i) with $a x^{2}+b x+c,$ we get,

a=2, b=-7, c=-1

$\because x=\frac{-b \pm \sqrt{b^{2}-4 a c}}{2 a}$

$\Rightarrow x=\frac{-(-7) \pm \sqrt{(-7)^{2}-4(2) \times(-1)}}{2 \times 2}$

$\Rightarrow \frac{7 \pm \sqrt{49+8}}{4}=\frac{7 \pm \sqrt{57}}{4}$

$\Rightarrow x=\frac{7+\sqrt{57}}{4}$ or $x=\frac{7-\sqrt{57}}{4}$

$\Rightarrow x=\frac{7+7.55}{4}$ or $x=\frac{7-7.55}{4}$

$\Rightarrow x=\frac{14.55}{4}$ or $x=\frac{-0.55}{4}$

$\Rightarrow x=3.64$ or $x=-0.14$ Ans.

Question 12

Solve the following equation: $x-\frac{18}{x}=6$. Give your answer correct to two x significant figures. (2011)

Sol :

$x-\frac{18}{x}=6$

⇒ x² – 6x – 18 = 0

a = 1, b = -6, c = -18

$x=\frac{-b \pm \sqrt{b^{2}-4 a c}}{2 a}=\frac{6 \pm \sqrt{36+72}}{2}$

$=\frac{6 \pm \sqrt{108}}{2}=\frac{6 \pm 6 \sqrt{3}}{2}=$ or $\frac{6(1-1.73)}{2}$

=3×2.73

or 3×-0.73=8.19 or -2.19

Question 13

Solve the equation 5x² – 3x – 4 = 0 and give your answer correct to 3 significant figures:

Sol :

We have 5x² – 3x – 4 = 0

Here a = 5, b = – 3, c = – 4

$x=\frac{-b \pm \sqrt{b^{2}-4 a c}}{2 a}=\frac{3 \pm \sqrt{9+4 \times 5 \times 4}}{2 \times 5}=\frac{3 \pm \sqrt{89}}{10}$

$x=\frac{3+9.43}{10}$ or $x=\frac{3-9.43}{10}$

$\Rightarrow x=\frac{12.43}{10},$ or $x=\frac{-6.43}{10}$

⇒x=1.24 or x=-0.643