ML Aggarwal Solution Class 10 Chapter 5 Quadratic Equations in One Variable Exercise 5.2
Exercise 5.2
Question 1
(i) 4x² = 3x
Sol :
(i) 4x² = 3x
x(4x – 3) = 0
Either x = 0,
$\therefore x=0, \frac{3}{4}$
(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
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(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
(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
Hence $x=\frac{5}{2}, \frac{-7}{3}$
Question 6
(i) 6p² + 11p – 10 = 0
Sol :
(i)
6p² + 11p – 10 = 0
⇒ 6p² + 15p – 4p – 10 = 0
⇒ 3p(2p + 5) – 2(2p + 5) = 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
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
$(3 x-5)^{2}=28 \times 7$
$9 x^{2}-30 x+25=196$
$9 x^{2}-30 x+25-196=0$
$9 x^{2}-30 x-171=0$
$3 x^{2}-10 x-57=0$ (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
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
(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
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
$a^{2} x+b^{2}=0,$ 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
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
Sol :
(i)
x² – (1 + √2)x + √2 = 0
⇒ x² – x – √2x + √2 = 0
(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
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
$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
⇒ 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
$p=\frac{1}{2}$
∴$p=\frac{1}{2},-3$
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