ML Aggarwal Solution Class 10 Chapter 5 Quadratic Equations in One Variable Exercise 5.2

Exercise 5.2

Question 1

(i) 4x² = 3x

(ii) $\frac{x^{2}-5 x}{2}=0$

Sol :

(i) 4x² = 3x

x(4x – 3) = 0

Either x = 0,

or 4x-3=0 then 4x=3

$x=\frac{3}{4}$

$\therefore x=0, \frac{3}{4}$

(ii) $\frac{x^{2}-5 x}{2}=0$

$x^{2}-5 x=0$

⇒x(x-5)=0

Either x=0 or x-5=0, then x=5

Hence , x=0,5

Question 2

(i) (x – 3) (2x + 5) = 0

(ii) x (2x + 1) = 6

Sol :

(i) 

(x – 3) (2x + 5) = 0

Either x – 3 = 0,

Then x = 3

2x+5=0

2x=-5

$x=\frac{-5}{2}$

Hence, $x=3,=\frac{-5}{2}$

(ii)

x(2x+1)=6

x(2 x+1)=6

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

$2 x^{2}+4 x-3 x-6=0$

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

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

Either x+2=0 , then x=-2

2x-3=0 then 

2x=3

$x=\frac{3}{2}$

Hence , $x=-2, \frac{3}{2}$

Question 3

(i) x² – 3x – 10 = 0

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

Sol :

(i) 

x² – 3x – 10 = 0

⇒ x² – 5x + 2x – 10 = 0

⇒ x(x – 5) + 2(x – 5) = 0

⇒ (x-5)(x+2)=0

Either x-5=0 then x=5

x+2=0 then x=-2

Hence , x=5,-2

(ii)

x(2x+5)=3

x(2x+5)=3

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

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

2 x(x+3)-1(x+3)=0

(x+3)(2 x-1)=0

Either x+3=0, then x=-3

2x-1=0 then 2x=1 

$x=\frac{1}{2}$

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

Question 4

(i) 3x² – 5x – 12 = 0

(ii) 21x² – 8x – 4 = 0

Sol :

(i) 3x² – 5x – 12 = 0

⇒ 3x² – 9x + 4x – 12 = 0

⇒ 3x (x – 3) + 4(x – 3) = 0

⇒ (x-3)(3x+4)=0

Either x-3=0, then x=3

3x+4=0 then 3x=-4

$x=\frac{-4}{3}$

Hence, $x=3, \frac{-4}{3}$

(ii)

$21 x^{2}-8 x-4=0$

$21 x^{2}-8 x-4=0$

$21 x^{2}-14 x+6 x-4=0$

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

(3 x-2)(7 x+2)=0

Either 3x-2=0 then 3x=2

$x=\frac{2}{3}$

or 7x+2=0 then 7x=-2

$x=\frac{-2}{7}$

Hence,$x=\frac{2}{3}, \frac{-2}{7}$ 

Question 5

(i) 3x² = x + 4

(ii) x(6x – 1) = 35

Sol :

(i) 

3x² = x + 4

⇒ 3x² – x – 4 = 0

⇒ 3x² – 4x + 3x – 4 = 0

⇒ x(3x-4)+1(3x-4)=0

⇒ (3x-4)(x+1)=0

Either 3x-4=0 , then 3x=4

$x=\frac{4}{3}$

or x+1=0 , then x=-1

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

(ii)

x(6 x-1)=35

x(6 x-1)=35

$6 x^{2}-x-35=0$

$6 x^{2}-15 x+14 x-35=0$

$3 x(2 x-5)+7(2 x-5)=0$

$(2 x-5)(3 x+7)=0$

Either 2x-5=0 then 2x=5

$x=\frac{5}{2}$

or 3x+7=0, then 3x=-7

$x=\frac{-7}{3}$

Hence $x=\frac{5}{2}, \frac{-7}{3}$

Question 6

(i) 6p² + 11p – 10 = 0

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

Sol :

(i) 

6p² + 11p – 10 = 0

⇒ 6p² + 15p – 4p – 10 = 0

⇒ 3p(2p + 5) – 2(2p + 5) = 0

⇒ (2 p+5)(3 p-2)=0

Either 2p+5=0, then 2 p=-5 $\Rightarrow p=\frac{-5}{2}$

or 3 p-2=0, then 3 p=2$\Rightarrow p=\frac{2}{3}$

Hence $p=\frac{-5}{2}, \frac{2}{3}$

(ii)

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

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

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

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

x(2 x-3)+1(2 x-3)=0

(2 x-3)(x+1)=0

Either 2 x-3=0, then 2 x=3 $\Rightarrow x=\frac{3}{2}$

x+1=0, then x=-1

x+1=0 then x=-1

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

Question 7

(i) (x – 4)² + 5² = 13²

(ii) 3(x – 2)² = 147

Sol :

(i) 

(x – 4)² + 5² = 13²

x² – 8x + 16 + 25 = 169

$x^{2}-8 x+41-169=0

$\Rightarrow x^{2}-8 x-128=0$

$\Rightarrow x^{2}-16 x+8 x-128=0$

x(x-16)+8(x-16)=0

(x-16)(x+8)=0

Either  x-16=0 then x=16

or x+8=0 then x=-8

Hence x=16,-8

(ii)

$3(x-2)^{2}=147$

$3(x-2)^{2}=147$

$3\left(x^{2}-4 x+4\right)=147$

$3 x^{2}-12 x+12-147=0$

$3 x^{2}-12 x-135=0$

$x^{2}-4 x-45=0$ (dividing by 3)

$x^{2}-9 x+5 x-45=0$

x(x-9)+5(x-9)=0

(x-9)(x+5)=0

Either x-9=0, then x=9

x+5=0 then x=-5

Hence x=9,-5

Question 8

(i) $\frac{1}{7}(3 x-5)^{2}=28$

(ii) 3(y² – 6) = y(y + 7) – 3

Sol :

(i) 

$\frac{1}{7}(3 x-5)^{2}=28$

$(3 x-5)^{2}=28 \times 7$

$9 x^{2}-30 x+25=196$

$9 x^{2}-30 x+25-196=0$

$9 x^{2}-30 x-171=0$

$3 x^{2}-10 x-57=0$ (dividing by 3)

$3 x^{2}-19 x+9 x-57=0$

x(3 x-19)+3(3 x-19)=0

(3 x-19)(x+3)=0

Either 3x-19=0 then 3x=19

$x=\frac{19}{3}$ or x+3=0, then x=-3

Hence $x=\frac{19}{3},-3$

(ii)

$3\left(y^{2}-6\right)=y(y+7)-3$

$3\left(y^{2}-6\right)=y(y+7)-3$

$3\left(y^{2}-6\right)=y^{2}+7 y-3$

$3 y^{2}-18=y^{2}+7 y-3$

$3 y^{2}-y^{2}-7 y-18+3=0$

$2 y^{2}-7 y-15=0$

$2 y^{2}-10 y+3 y-15=0$

2y(y-5)+3(y-5)=0

(y-5)(2 y+3)=0

Either y-5=0, then y=5

or 2y+3=0 then 2y=-3 $\Rightarrow y=\frac{-3}{2}$

Hence $y=\frac{-3}{2},5$

Question 9

x² – 4x – 12 = 0,when x∈N

Sol :

x² – 4x – 12 = 0

⇒ x² – 6x + 2x – 12 = 0

⇒ x (x – 6) + 2 (x – 6) = 0

⇒ (x – 6) (x + 2) = 0

Either x – 6 = 0, then x = 6

or x + 2 = 0, then x = -2

But -2 is not a natural number

∴ x = 6

Question 10

2x² – 8x – 24 = 0 when x∈I

Sol :

2x² – 8x – 24 = 0

⇒ x² – 4x – 12 = 0 (Dividing by 2)

⇒ x² – 6x + 2x – 12 = 0

⇒ x (x – 6) + 2 (x – 6) = 0

⇒ (x – 6) (x + 2) = 0

Either x – 6 = 0, then, x = 6

or x + 2 = 0, then x = – 2

Hence x = 6, – 2

Question 11

5x² – 8x – 4 = 0 when x∈Q

Sol :

5x² – 8x – 4 = 0

∵ 5 × ( – 4) = – 20

-20 = – 10 + 2

-8 = – 10 + 2

$5 x^{2}-10 x+2 x-4=0$

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

(x-2)(5 x+2)=0 [zero product rule]

Either x-2=0, then x=2

5x+2=0 then 5x=-2

$x=-\frac{2}{5}$

$\therefore, x=2,-\frac{2}{5}$

Question 12

2x² – 9x + 10 = 0,when

(i)x∈N

(ii)x∈Q

Sol :

2x² – 9x + 10 = 0

⇒ 2x² – 4x – 5x + 10 = 0

⇒ 2x(x – 2) – 5(x – 2) = 0

(x-2)(2 x-5)=0

Either x-2=0 then x=2

or 2x-5=0 then 2 x=5 $\Rightarrow x=\frac{5}{2}$

(i) when x∈N , then x=2

(ii) when x∈Q, then $x=2, \frac{5}{2}$

Question 13

(i) a²x² + 2ax + 1 = 0, a≠0

(ii) x² – (p + q)x + pq = 0

Sol :

(i)

 a²x² + 2ax + 1 = 0

⇒ a²x² + ax + ax + 1 = 0

$\Rightarrow a x(a x+1)+1(a x+1)=0$

$\Rightarrow(a x+1)(a x+1)=0$

$\Rightarrow(a x+1)^{2}=0$

∴ax+1=0 or ax=-1

$x=-\frac{1}{a}$

Hence $x=-\frac{1}{a},-\frac{1}{a}$

(ii)

$x^{2}-(p+q) x+p q=0$

$x^{2}-(p+q) x+p q=0$

$x^{2}-p x-q x+p q=0$

x(x-p)-q(x-p)=0

(x-p)(x-q)=0

Either x-p=0 then x=p

x-q=0 then x=q

Hence x=p, q

Question 14

a²x² + (a² + b²)x + b² = 0, a≠0

Sol :

a²x² + (a² + b²)x + b² = 0

⇒ a²x(x + 1) + b²(x + 1) = 0

$(x+1)\left(a^{2} x+b^{2}\right)=0$

(x+1)=0, then x=-1

$a^{2} x+b^{2}=0,$ then $a^{2} x=-b^{2} \quad \Rightarrow x=\frac{-b^{2}}{a^{2}}$

Hence $x=-1, \frac{-b^{2}}{a^{2}}$

Question 15

(i) √3x² + 10x + 7√3 = 0

(ii) 4√3x² + 5x – 2√3 = 0

Sol :

(i)

√3x² + 10x + 7√3 = 0

[ ∵ √3 x 7√3 = 7 x 3 = 21]

⇒ √3x(x + √3) + 7(x + √3) = 0

⇒ $(x+\sqrt{3})(\sqrt{3} x+7)=0$

Either $x+\sqrt{3}=0,$ then $x=-\sqrt{3}$

or $\sqrt{3} x+7=0,$ then $\sqrt{3} x=-7$

$x=\frac{-7}{\sqrt{3}}$

$\Rightarrow x=\frac{-7 \times \sqrt{3}}{\sqrt{3} \times \sqrt{3}}$

$=\frac{-7 \sqrt{3}}{3}$

Hence $x=-\sqrt{3},-\frac{7 \sqrt{3}}{3}$

(ii)

⇒$4 \sqrt{3} x^{2}+5 x-2 \sqrt{3}=0$

⇒$4 \sqrt{3} x^{2}+5 x-2 \sqrt{3}=0$

⇒$\{4 \sqrt{3} \times(-2 \sqrt{3})=8 \times(-3)=-24\}$

⇒$4 \sqrt{3} x^{2}+8 x-3 x-2 \sqrt{3}=0$

⇒$4 x(\sqrt{3} x+2)-\sqrt{3}(\sqrt{3} x+2)=0$

⇒$(\sqrt{3} x+2)(4 x-\sqrt{3})=0$

Either $\sqrt{3} x+2=0,$ then $\sqrt{3} x=-2$

⇒$x=-\frac{2}{\sqrt{3}}$

⇒$x=-\frac{-2 \times \sqrt{3}}{\sqrt{3} \times \sqrt{3}}=-\frac{-2 \sqrt{3}}{3}$

⇒$4 x-\sqrt{3}=0,$ then $4 x=\sqrt{3}$

⇒$x=\frac{\sqrt{3}}{4}$

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

Question 16

(i) x² – (1 + √2)x + √2 = 0

(ii) $x+\frac{1}{x}=2 \frac{1}{20}$

Sol :

(i) 

x² – (1 + √2)x + √2 = 0

⇒ x² – x – √2x + √2 = 0

$x(x-1)-\sqrt{2}(x-1)=0$

$(x-1)(x-\sqrt{2})=0$

Either $x-1=0,$ then $x=1$

or $x-\sqrt{2}=0,$ then $x=\sqrt{2}$

Hence $x=1, \sqrt{2}$

(ii)

$x+\frac{1}{x}=2 \frac{1}{20}$

$\frac{x^{2}+1}{x}=\frac{41}{20}$

$20 x^{2}+20=41 x$

$20 x^{2}-16 x-25 x+20=0$

4 x(5 x-4)-5(5 x-4)=0

(5 x-4)(4 x-5)=0

Either 5x-4=0, then 5x=4$\Rightarrow x=\frac{4}{5}$

or 4x-5=0, then 4x=5 $\Rightarrow x=\frac{5}{4}$

Hence $x=\frac{4}{5}, \frac{5}{4}$ Ans.

Question 17

(i) $\frac{2}{x^{2}}-\frac{5}{x}+2=0, x \neq 0$

(ii) $\frac{x^{2}}{15}-\frac{x}{3}-10=0$

Sol :

(i)

$\frac{2}{x^{2}}-\frac{5}{2}+2=0, x \neq 0$

⇒ 2 – 5x + 2x² = 0

$\Rightarrow 2 x^{2}-5 x+2=0$ 

$\left\{\begin{array}{c}\because 2 \times 2=4 \\ 4=-4 \times(-1) \\ -5=-4-1\end{array}\right\}$

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

⇒2x(x-2)-1(x-2)=0

⇒(x-2)(2 x-1)=0

Either x-2=0, then x=2

or 2 x-1=0, then 2 x=1 $\Rightarrow x=\frac{1}{2}$

$\therefore x=2, \frac{1}{2}$

(ii)

⇒$\frac{x^{2}}{15}-\frac{x}{3}-10=0$

⇒$x^{2}-5 x-150=0$ $\left\{\begin{array}{c}\because-150=-15 \times 10 \\ -5=-15+10\end{array}\right\}$

⇒$x^{2}-15 x+10 x-150=0$

⇒x(x-15)+10(x-15)=0

⇒(x-15)(x+10)=0

Either x-15=0, then x=15

⇒or x+10=0, then x=-10

⇒x=15,-10

Question 18

(i) $3 x-\frac{8}{x}=2$

(ii) $\frac{x+2}{x+3}=\frac{2 x-3}{3 x-7}$

Sol :

(i)

⇒$3 x-\frac{8}{x}=2$

⇒$\frac{3 x^{2}-8}{x}=2$

⇒$3 x^{2}-8=2 x$

⇒$3 x^{2}-2 x-8=0$

⇒$3 x^{2}-6 x+4 x-8=0$

⇒3x(x-2)+4(x-2)=0

⇒(x-2)(3 x+4)=0

Either x-2=0, then x=2

⇒or  x+4=0,$then 3x=-4$\Rightarrow x=\frac{-4}{3}$

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

(ii)

⇒$\frac{x+2}{x+3}=\frac{2 x-3}{3 x-7}$

⇒(x+2)(3 x-7)=(2 x-3)(x+3)

$\Rightarrow 3 x^{2}-7 x+6 x-14=2 x^{2}+6 x-3 x-9$

$\Rightarrow 3 x^{2}-x-14=2 x^{2}+3 x-9$

$\Rightarrow 3 x^{2}-x-14-2 x^{2}-3 x+9=0$

$\Rightarrow x^{2}-4 x-5=0$

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

⇒x(x-5)+1(x-5)=0

⇒(x-5)(x+1)=0

Either x-5=0, then x=5

⇒or x+1=0, then x=-1

Hence x=5,-1

Question 19

(i) $\frac{8}{x+3}-\frac{3}{2-x}=2$

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

Sol :

(i)

$\frac{8}{x+3}-\frac{3}{2-x}=2$

$\frac{16-8 x-3 x-9}{(x+3)(2-x)}=2$

$\Rightarrow \frac{-11 x+7}{2 x-x^{2}+6-3 x}=2$

$\Rightarrow-11 x+7=4 x-2 x^{2}+12-6 x$

$\Rightarrow-11 x+7-4 x+2 x^{2}-12+6 x=0$

$\Rightarrow 2 x^{2}-9 x-5=0$

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

⇒2x(x-5)+1(x-5)=0

⇒(x-5)(2 x+1)=0

Either x-5=0,$then$x=5$

or 2 x+1=0, then 2 x=-1 

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

Hence $x=5,-\frac{1}{2}$ Ans.

(ii)

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

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

$\frac{x}{x-1}+\frac{x-1}{x}=\frac{5}{2}$

$\frac{x^{2}+x^{2}-2 x+1}{x(x-1)}=\frac{5}{2}$

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

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

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

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

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

⇒x(x-2)+1(x-2)=0

⇒(x-2)(x+1)=0

Either x-2=0, then x=2

or x+1=0, then x=-1

Hence x=2,-1

Question 20

(i) $\frac{x}{x+1}+\frac{x+1}{x}=\frac{34}{15}$

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

Sol :

(i)

$\frac{x}{x+1}+\frac{x+1}{x}=\frac{34}{15}$

$\frac{x^{2}+x^{2}+2 x+1}{x(x+1)}=\frac{34}{15}$

$\Rightarrow \quad \frac{2 x^{2}+2 x+1}{x^{2}+x}=\frac{34}{15}$

$\Rightarrow 30 x^{2}+30 x+15=34 x^{2}+34 x$

$\Rightarrow 30 x^{2}+30 x+15-34 x^{2}-34 x=0$

$\Rightarrow-4 x^{2}-4 x+15=0$

$\Rightarrow 4 x^{2}+4 x-15=0$

$\Rightarrow 4 x^{2}+10 x-6 x-15=0$

⇒2x(2 x+5)-3(2x+5)=0

⇒(2x+5)(2 x-3)=0

Either 2x+5=0, then 2x=-5

 $\Rightarrow x=\frac{-5}{2}$

or 2x-3=0, then 2x=3$\Rightarrow x=\frac{3}{2}$

Hence $x=\frac{-5}{2}, \frac{3}{2}$ Ans.

(ii)

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

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

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

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

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

$\Rightarrow 2 x^{2}+4=3 x^{2}+3 x-6$

$\Rightarrow 2 x^{2}+4-3 x^{2}-3 x+6=0$

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

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

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

⇒x(x+5)-2(x+5)=0

⇒(x+5)(x-2)=0

 Either x+5=0, then x=-5

or x-2=0, then x=2

Hence x=-5,2 Ans.

Question 21

(i) $\frac{1}{x-3}-\frac{1}{x+5}=\frac{1}{6}$

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

Sol :

(i)

$\frac{1}{x-3}-\frac{1}{x+5}=\frac{1}{6}$

$\frac{x+5-x+3}{(x-3)(x+5)}=\frac{1}{6} \Rightarrow \frac{8}{x^{2}+2 x-15}=\frac{1}{6}$

$\Rightarrow x^{2}+2 x-15=48$

$\Rightarrow x^{2}+2 x-15-48=0$

$\Rightarrow x^{2}+2 x-63=0 \Rightarrow x^{2}+9 x-7 x-63=0$

⇒x(x+9)-7(x+9)=0

⇒(x+9)(x-7)=0

Either x+9=0, then x=-9

or x-7=0, then x=7

Hence x=-9,7

(ii)

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

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

Put $\frac{x-3}{x+3}=a,$ then $\frac{x+3}{x-3}=\frac{1}{a}$

$\therefore a+\frac{1}{a}=\frac{5}{2}$

$2 a^{2}+2=5 a$

$\Rightarrow 2 a^{2}-5 a+2=0 \Rightarrow 2 a^{2}-a-4 a+2=0$

⇒a(2a-1)-2(2a-1)=0

⇒2a-1)(a-2)=0

Either 2a-1=0, then $a=\frac{1}{2}$

or a-2=0, then a=2

(a) When $a=\frac{1}{2},$ then

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

⇒2x-6=x+3

2x-x=3+6

x=9

(b)When $a=2,$ then

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

2x+6=x-3

2x-x=-3-6

x=-9

∴x=9,-9

Question 22

(i) $\frac{a}{a x-1}+\frac{b}{b x-1}=a+b, a+b \neq 0, a b \neq 0$

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

Sol :

(i) $\frac{a}{a x-1}+\frac{b}{b x-1}=a+b$

$\Rightarrow \left(\frac{a}{a x-1}-b\right)+\left(\frac{b}{b x-1}-a\right)=0$

$\Rightarrow \frac{a-a b x+b}{(a x-1)}+\frac{b-a b x+a}{(b x-1)}=0$

$\Rightarrow(a+b-a b x)\left[\frac{1}{a x-1}+\frac{1}{b x-1}\right]=0$

$\Rightarrow (a+b-a b x)\left[\frac{b x-1+a x-1}{(a x-1)(b x-1)}\right]=0$

$\Rightarrow \frac{(a+b-a b x)(a x+b x-2)}{(a x-1)(b x-1)}=0$

$\Rightarrow(a+b-a b x)(a x+b x-2)=0$

Either $a+b-a b x=0,$ then $a+b=a b x$

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

or ax+bx-2=0, then x(a+b)=2

$x=\frac{2}{a+b}$

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

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

$\Rightarrow \quad \frac{1}{2 a+b+2 x}-\frac{1}{2 x}=\frac{1}{2 a}+\frac{1}{b}$

$\Rightarrow \frac{2 x-(2 a+b+2 x)}{(2 a+b+2 x) 2 x}=\frac{b+2 a}{2 a b}$

$\Rightarrow \frac{-(2 a+b)}{(2 a+b+2 x) 2 x}=\frac{(2 a+b)}{2 a b}$

$\Rightarrow \frac{-1}{(2 a+b+2 x) 2 x}=\frac{1}{2 a b}$

⇒-2ab=(2a+b+2x)2x

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

$\Rightarrow 4 x^{2}+2 b x+4 a x+2 a b=0$

⇒2 x(2 x+b)+2a(2 x+b)=0

⇒(2 x+2 a)(2 x+b)=0

⇒2x+2a=0 or 2x+b=0

x=-a or $x=\frac{-b}{2}$

Hence, the roots of the given equation are

-a and $\frac{-b}{2} .$ Ans.

Question 23

$\frac{1}{x+6}+\frac{1}{x-10}=\frac{3}{x-4}$

Sol :

$\frac{1}{x+6}+\frac{1}{x-10}=\frac{3}{x-4}$

$\frac{x-10+x+6}{(x+6)(x-10)}=\frac{3}{x-4}$

$\Rightarrow 2 x^{2}-8 x-4 x+16=3\left(x^{2}-4 x-60\right)$

$\Rightarrow 2 x^{2}-8 x-4 x+16=3 x^{2}-12 x-180$

$\Rightarrow 2 x^{2}-12 x+16-3 x^{2}+12 x+180=0$

$\Rightarrow-x^{2}+196=0$

$\Rightarrow x^{2}-196=0$

$\Rightarrow (x)^{2}-(14)^{2}=0$

⇒(x+14)(x-14)=0

Either x+14=0, then x=-14

or x-14=0, then x=14

∴ x=14,-14 

Question 24

(i) $\sqrt{3 x+4}=x$

(ii) $\sqrt{x(x-7)}=3 \sqrt{2}$

Sol :

(i) $\sqrt{3 x+4}=x$

Squaring on both sides

$3 x+4=x^{2}$

$\Rightarrow x^{2}-3 x-4=0$

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

⇒x(x-4)+1(x-4)=0

⇒(x-4)(x+1)=0

Either x-4=0, then x=4 or x+1=0, then x=-1

∴ x=4,-1

Check (i) If x=4, then L.H.S. $=\sqrt{3 x+4}=\sqrt{3 \times 4+4}$

$=\sqrt{12+4}=\sqrt{16}=4$

R.H.S. =x=4 

∴ L.H.S=R.H.S

Hence x=4 is its root

(ii) If x=-1, then

L.H.S. $=\sqrt{3 \times(-1)+4}=\sqrt{-3+4}=\sqrt{1}=1$

R.H.S.=x=-1

∵ L.H.S.≠R.H.S

∴ x=-1 is not its root, Hence x=4

(ii)

$\sqrt{x(x-7)}=3 \sqrt{2}$

$\sqrt{x(x-7)}=3 \sqrt{2},$ 

Squaring both sides, 

x(x-7)=9×2

$\Rightarrow x^{2}-7 x=18$

$\Rightarrow  x^{2}-7 x-18=0$

$\Rightarrow x^{2}-9 x+2 x-18=0$

⇒x(x-9)+2(x-9)=0

⇒(x-9)(x+2)=0

Either x-9=0, then x=9 or  x+2=0, then x=-2

∴ x=9,-2

Check :

(i) If x=9, then L.H.S. $=\sqrt{x(x-7)}=\sqrt{9(9-7)}$

$=\sqrt{9 \times 2}=\sqrt{18}=\sqrt{9 \times 2}=3 \sqrt{2}=\mathrm{R.H.S}$

x=9 is a root

(ii) If x=-2 then

L.H.S. $=\sqrt{x(x-7)}=\sqrt{-2(-2-7)}$

$=\sqrt{-2 \times-9}=\sqrt{18}=\sqrt{9 \times 2}=3 \sqrt{2}=\mathrm{R.H.S}$

∴x=-2 is also its root Hence x=9,-2

Question 25

Use the substitution y = 3x + 1 to solve for x : 5(3x + 1 )² + 6(3x + 1) – 8 = 0

Sol :

y = 3x + 1

Now, 5(3x + 1)² + 6(3x + 1) – 8 = 0

Substituting the value of  3x+1, we get

$5 y^{2}+6 y-8=0$

$\Rightarrow 5 y^{2}+10 y-4 y-8=0$

$\left\{\begin{array}{c}\because 5 \times(-8)=-40 \\ \therefore-40=10 \times(-4) \\ 6=10-4\end{array}\right\}$

⇒5y(y+2)-4(y+2)=0

⇒(y+2)(5y-4)=0

Either y+2=0, then y=-2

or 5y-4=0 then 5y=4 $\Rightarrow y=\frac{4}{5}$

(i) If y=-2, then 3x+1=-2 $\Rightarrow 3 x=-2-1$

⇒3x=-3

$\Rightarrow x=\frac{-3}{3}=-1$

(ii)

If $y=\frac{4}{5},$ then $3 x+1=\frac{4}{5}$

$\Rightarrow 3 x=\frac{4}{5}-1=\frac{-1}{5}$

Hence $x=-1, \frac{-1}{15}$

Question 26

Find the values of x if p + 1 =0 and x² + px – 6 = 0

Sol :

p + 1 = 0, then p = – 1

Substituting the value of p in the given quadratic equation

x² + ( – 1)x – 6 = 0

⇒ x² – x – 6 = 0

⇒ x² – 3x + 2x – 6 = 0

⇒ x (x – 3) + 2 (x – 3) = 0

⇒ (x – 3) (x + 2) = 0

Either x – 3 = 0, then x = 3

or x + 2 = 0, then x = – 2

Hence x = 3, -2

Question 27

Find the values of x if p + 7 = 0, q – 12 = 0 and x² + px + q = 0,

Sol :

p + 7 = 0, then p = – 7

and q – 12 = 0, then q = 12

Substituting the values of p and q in the given quadratic equation,

x² – 7x + 12 = 0

⇒ x² – 3x – 4x + 12 = 0

⇒ x (x – 3) – 4 (x – 3) = 0

⇒ (x – 3) (x – 4) = 0

Either x – 3 = 0, then x = 3

or x – 4 = 0, then x = 4

Hence x = 3, 4

Question 29

If x = p is a solution of the equation x(2x + 5) = 3, then find the value of p.

Sol :

Given, x = p and x(2x + 5) = 3

Substituting the value of p, we get

p(2p + 5) = 3

⇒ 2p² + 5p – 3 = 0

⇒ 2p² + 6p – p – 3 = 0

$\left\{\begin{array}{c}\because 2 \times(-3)=-6 \\ \therefore-6=6 \times(-1) \\ 5=6-1\end{array}\right\}$

⇒ 2p(p+3)-1(p+3)=9

⇒ (p+3)(2 p-1)=0

Either p+3=0 then p=-3

or 2p-1=0 then 2p=1

$p=\frac{1}{2}$

∴$p=\frac{1}{2},-3$