Class 10: Quadratic Equations

$\displaystyle \textbf{Question 1: }\text{Solve the following quadratic equation by factorization:}$
$\displaystyle 25x(x+1)=-4$
$\displaystyle \text{Answer:}$
$\displaystyle 25x(x+1)=-4$
$\displaystyle \Rightarrow 25x^2+25x+4=0$
$\displaystyle \Rightarrow 25x^2+20x+5x+4=0$
$\displaystyle \Rightarrow 5x(5x+4)+1(5x+4)=0$
$\displaystyle \Rightarrow (5x+4)(5x+1)=0$
$\displaystyle \Rightarrow 5x+4=0\text{ or }5x+1=0$
$\displaystyle \therefore x=-\frac{4}{5}\text{ or }x=-\frac{1}{5}$
$\displaystyle \\$

$\displaystyle \textbf{Question 2: }\text{Solve the following quadratic equation by factorization:}$
$\displaystyle 16x-\frac{10}{x}=27,\quad x\ne0\hfill\text{[CBSE 2014]}$
$\displaystyle \text{Answer:}$
$\displaystyle 16x-\frac{10}{x}=27$
$\displaystyle \Rightarrow 16x^2-10=27x$
$\displaystyle \Rightarrow 16x^2-27x-10=0$
$\displaystyle \Rightarrow 16x^2-32x+5x-10=0$
$\displaystyle \Rightarrow 16x(x-2)+5(x-2)=0$
$\displaystyle \Rightarrow (x-2)(16x+5)=0$
$\displaystyle \Rightarrow x-2=0\text{ or }16x+5=0$
$\displaystyle \therefore x=2\text{ or }x=-\frac{5}{16}$
$\displaystyle \\$

$\displaystyle \textbf{Question 3: }\text{Solve the following quadratic equation by factorization:}$
$\displaystyle 6x^2+11x+3=0\hfill\text{[CBSE 2010]}$
$\displaystyle \text{Answer:}$
$\displaystyle 6x^2+11x+3=0$
$\displaystyle \Rightarrow 6x^2+9x+2x+3=0$
$\displaystyle \Rightarrow 3x(2x+3)+1(2x+3)=0$
$\displaystyle \Rightarrow (2x+3)(3x+1)=0$
$\displaystyle \Rightarrow 2x+3=0\text{ or }3x+1=0$
$\displaystyle \therefore x=-\frac{3}{2}\text{ or }x=-\frac{1}{3}$
$\displaystyle \\$

$\displaystyle \textbf{Question 4: }\text{Solve the following quadratic equation by factorization:}$
$\displaystyle 2x^2+ax-a^2=0\hfill\text{[CBSE 2014]}$
$\displaystyle \text{Answer:}$
$\displaystyle 2x^2+ax-a^2=0$
$\displaystyle \Rightarrow 2x^2+2ax-ax-a^2=0$
$\displaystyle \Rightarrow 2x(x+a)-a(x+a)=0$
$\displaystyle \Rightarrow (x+a)(2x-a)=0$
$\displaystyle \Rightarrow x+a=0\text{ or }2x-a=0$
$\displaystyle \therefore x=-a\text{ or }x=\frac{a}{2}$
$\displaystyle \\$

$\displaystyle \textbf{Question 5: }\text{Solve the following quadratic equation by factorization:}$
$\displaystyle \frac{1}{x-1}-\frac{1}{x+5}=\frac{6}{7},\quad x\ne1,-5\hfill\text{[CBSE 2010]}$
$\displaystyle \text{Answer:}$
$\displaystyle \frac{1}{x-1}-\frac{1}{x+5}=\frac{6}{7}$
$\displaystyle \Rightarrow \frac{(x+5)-(x-1)}{(x-1)(x+5)}=\frac{6}{7}$
$\displaystyle \Rightarrow \frac{6}{(x-1)(x+5)}=\frac{6}{7}$
$\displaystyle \Rightarrow (x-1)(x+5)=7$
$\displaystyle \Rightarrow x^2+4x-5=7$
$\displaystyle \Rightarrow x^2+4x-12=0$
$\displaystyle \Rightarrow x^2+6x-2x-12=0$
$\displaystyle \Rightarrow x(x+6)-2(x+6)=0$
$\displaystyle \Rightarrow (x+6)(x-2)=0$
$\displaystyle \Rightarrow x+6=0\text{ or }x-2=0$
$\displaystyle \therefore x=-6\text{ or }x=2$
$\displaystyle \\$

$\displaystyle \textbf{Question 6: }\text{Solve the following quadratic equation by factorization:}$
$\displaystyle \frac{1}{x+4}-\frac{1}{x-7}=\frac{11}{30},\quad x\ne-4,7$
$\displaystyle \text{Answer:}$
$\displaystyle \frac{1}{x+4}-\frac{1}{x-7}=\frac{11}{30}$
$\displaystyle \Rightarrow \frac{(x-7)-(x+4)}{(x+4)(x-7)}=\frac{11}{30}$
$\displaystyle \Rightarrow \frac{-11}{(x+4)(x-7)}=\frac{11}{30}$
$\displaystyle \Rightarrow (x+4)(x-7)=-30$
$\displaystyle \Rightarrow x^2-3x-28=-30$
$\displaystyle \Rightarrow x^2-3x+2=0$
$\displaystyle \Rightarrow x^2-2x-x+2=0$
$\displaystyle \Rightarrow x(x-2)-1(x-2)=0$
$\displaystyle \Rightarrow (x-1)(x-2)=0$
$\displaystyle \Rightarrow x-1=0\text{ or }x-2=0$
$\displaystyle \therefore x=1\text{ or }x=2$
$\displaystyle \\$

$\displaystyle \textbf{Question 7: }\text{Solve the following quadratic equation by factorization:}$
$\displaystyle \frac{1}{x-3}+\frac{2}{x-2}=\frac{8}{x},\quad x\ne0,2,3\hfill\text{[CBSE 2013]}$
$\displaystyle \text{Answer:}$
$\displaystyle \frac{1}{x-3}+\frac{2}{x-2}=\frac{8}{x}$
$\displaystyle \Rightarrow x(x-2)+2x(x-3)=8(x-3)(x-2)$
$\displaystyle \Rightarrow x^2-2x+2x^2-6x=8(x^2-5x+6)$
$\displaystyle \Rightarrow 3x^2-8x=8x^2-40x+48$
$\displaystyle \Rightarrow 5x^2-32x+48=0$
$\displaystyle \Rightarrow 5x^2-20x-12x+48=0$
$\displaystyle \Rightarrow 5x(x-4)-12(x-4)=0$
$\displaystyle \Rightarrow (x-4)(5x-12)=0$
$\displaystyle \Rightarrow x-4=0\text{ or }5x-12=0$
$\displaystyle \therefore x=4\text{ or }x=\frac{12}{5}$
$\displaystyle \\$

$\displaystyle \textbf{Question 8: }\text{Solve the following quadratic equation by factorization:}$
$\displaystyle \frac{16}{x}-1=\frac{15}{x+1},\quad x\ne0,-1\hfill\text{[CBSE 2014]}$
$\displaystyle \text{Answer:}$
$\displaystyle \frac{16}{x}-1=\frac{15}{x+1}$
$\displaystyle \Rightarrow 16(x+1)-x(x+1)=15x$
$\displaystyle \Rightarrow 16x+16-x^2-x=15x$
$\displaystyle \Rightarrow x^2-16=0$
$\displaystyle \Rightarrow (x-4)(x+4)=0$
$\displaystyle \Rightarrow x-4=0\text{ or }x+4=0$
$\displaystyle \therefore x=4\text{ or }x=-4$
$\displaystyle \\$

$\displaystyle \textbf{Question 9: }\text{Solve the following quadratic equation by factorization:}$
$\displaystyle \frac{x+3}{x+2}=\frac{3x-7}{2x-3},\quad x\ne-2,\frac{3}{2}\hfill\text{[CBSE 2017]}$
$\displaystyle \text{Answer:}$
$\displaystyle \frac{x+3}{x+2}=\frac{3x-7}{2x-3}$
$\displaystyle \Rightarrow (x+3)(2x-3)=(3x-7)(x+2)$
$\displaystyle \Rightarrow 2x^2+3x-9=3x^2-x-14$
$\displaystyle \Rightarrow x^2-4x-5=0$
$\displaystyle \Rightarrow x^2-5x+x-5=0$
$\displaystyle \Rightarrow x(x-5)+1(x-5)=0$
$\displaystyle \Rightarrow (x-5)(x+1)=0$
$\displaystyle \Rightarrow x-5=0\text{ or }x+1=0$
$\displaystyle \therefore x=5\text{ or }x=-1$
$\displaystyle \\$

$\displaystyle \textbf{Question 10: }\text{Solve the following quadratic equation by factorization:}$
$\displaystyle \frac{x+3}{x-2}-\frac{1-x}{x}=\frac{17}{4},\quad x\ne0,2\hfill\text{[CBSE 2017]}$
$\displaystyle \text{Answer:}$
$\displaystyle \frac{x+3}{x-2}-\frac{1-x}{x}=\frac{17}{4}$
$\displaystyle \Rightarrow \frac{x(x+3)-(1-x)(x-2)}{x(x-2)}=\frac{17}{4}$
$\displaystyle \Rightarrow \frac{x^2+3x-(x-2-x^2+2x)}{x^2-2x}=\frac{17}{4}$
$\displaystyle \Rightarrow \frac{2x^2+2}{x^2-2x}=\frac{17}{4}$
$\displaystyle \Rightarrow 8x^2+8=17x^2-34x$
$\displaystyle \Rightarrow 9x^2-34x-8=0$
$\displaystyle \Rightarrow 9x^2-36x+2x-8=0$
$\displaystyle \Rightarrow 9x(x-4)+2(x-4)=0$
$\displaystyle \Rightarrow (x-4)(9x+2)=0$
$\displaystyle \Rightarrow x-4=0\text{ or }9x+2=0$
$\displaystyle \therefore x=4\text{ or }x=-\frac{2}{9}$
$\displaystyle \\$

$\displaystyle \textbf{Question 11: }\text{Solve the following quadratic equation by factorization:}$
$\displaystyle \frac{x-3}{x+3}-\frac{x+3}{x-3}=\frac{48}{7},\quad x\ne3,-3\hfill\text{[CBSE 2015]}$
$\displaystyle \text{Answer:}$
$\displaystyle \frac{x-3}{x+3}-\frac{x+3}{x-3}=\frac{48}{7}$
$\displaystyle \Rightarrow \frac{(x-3)^2-(x+3)^2}{(x+3)(x-3)}=\frac{48}{7}$
$\displaystyle \Rightarrow \frac{x^2-6x+9-(x^2+6x+9)}{x^2-9}=\frac{48}{7}$
$\displaystyle \Rightarrow \frac{-12x}{x^2-9}=\frac{48}{7}$
$\displaystyle \Rightarrow -84x=48(x^2-9)$
$\displaystyle \Rightarrow 48x^2+84x-432=0$
$\displaystyle \Rightarrow 4x^2+7x-36=0$
$\displaystyle \Rightarrow 4x^2+16x-9x-36=0$
$\displaystyle \Rightarrow 4x(x+4)-9(x+4)=0$
$\displaystyle \Rightarrow (x+4)(4x-9)=0$
$\displaystyle \Rightarrow x+4=0\text{ or }4x-9=0$
$\displaystyle \therefore x=-4\text{ or }x=\frac{9}{4}$
$\displaystyle \\$

$\displaystyle \textbf{Question 12: }\text{Solve the following quadratic equation by factorization:}$
$\displaystyle \frac{4}{x}-3=\frac{5}{2x+3},\quad x\ne0,-\frac{3}{2}\hfill\text{[CBSE 2013]}$
$\displaystyle \text{Answer:}$
$\displaystyle \frac{4}{x}-3=\frac{5}{2x+3}$
$\displaystyle \Rightarrow \frac{4-3x}{x}=\frac{5}{2x+3}$
$\displaystyle \Rightarrow (4-3x)(2x+3)=5x$
$\displaystyle \Rightarrow 8x+12-6x^2-9x=5x$
$\displaystyle \Rightarrow 6x^2+6x-12=0$
$\displaystyle \Rightarrow x^2+x-2=0$
$\displaystyle \Rightarrow x^2+2x-x-2=0$
$\displaystyle \Rightarrow x(x+2)-1(x+2)=0$
$\displaystyle \Rightarrow (x+2)(x-1)=0$
$\displaystyle \Rightarrow x+2=0\text{ or }x-1=0$
$\displaystyle \therefore x=-2\text{ or }x=1$
$\displaystyle \\$

$\displaystyle \textbf{Question 13: }\text{Solve the following quadratic equation by factorization:}$
$\displaystyle \frac{3}{x+1}-\frac{1}{2}=\frac{2}{3x-1},\quad x\ne-1,\frac{1}{3}\hfill\text{[CBSE 2014, 2024]}$
$\displaystyle \text{Answer:}$
$\displaystyle \frac{3}{x+1}-\frac{1}{2}=\frac{2}{3x-1}$
$\displaystyle \Rightarrow 6(3x-1)-(x+1)(3x-1)=4(x+1)$
$\displaystyle \Rightarrow 18x-6-(3x^2+2x-1)=4x+4$
$\displaystyle \Rightarrow -3x^2+16x-5=4x+4$
$\displaystyle \Rightarrow 3x^2-12x+9=0$
$\displaystyle \Rightarrow x^2-4x+3=0$
$\displaystyle \Rightarrow x^2-3x-x+3=0$
$\displaystyle \Rightarrow x(x-3)-1(x-3)=0$
$\displaystyle \Rightarrow (x-3)(x-1)=0$
$\displaystyle \Rightarrow x-3=0\text{ or }x-1=0$
$\displaystyle \therefore x=3\text{ or }x=1$
$\displaystyle \\$

$\displaystyle \textbf{Question 14: }\text{Solve the following quadratic equation by factorization:}$
$\displaystyle \frac{3}{x+1}+\frac{4}{x-1}=\frac{29}{4x-1},\quad x\ne1,-1,\frac{1}{4}\hfill\text{[CBSE 2015]}$
$\displaystyle \text{Answer:}$
$\displaystyle \frac{3}{x+1}+\frac{4}{x-1}=\frac{29}{4x-1}$
$\displaystyle \Rightarrow \frac{3(x-1)+4(x+1)}{(x+1)(x-1)}=\frac{29}{4x-1}$
$\displaystyle \Rightarrow \frac{7x+1}{x^2-1}=\frac{29}{4x-1}$
$\displaystyle \Rightarrow (7x+1)(4x-1)=29(x^2-1)$
$\displaystyle \Rightarrow 28x^2-3x-1=29x^2-29$
$\displaystyle \Rightarrow x^2+3x-28=0$
$\displaystyle \Rightarrow x^2+7x-4x-28=0$
$\displaystyle \Rightarrow x(x+7)-4(x+7)=0$
$\displaystyle \Rightarrow (x+7)(x-4)=0$
$\displaystyle \Rightarrow x+7=0\text{ or }x-4=0$
$\displaystyle \therefore x=-7\text{ or }x=4$
$\displaystyle \\$

$\displaystyle \textbf{Question 15: }\text{Solve the following quadratic equation by factorization:}$
$\displaystyle \frac{2}{x+1}+\frac{3}{2(x-2)}=\frac{23}{5x},\quad x\ne0,-1,2\hfill\text{[CBSE 2015]}$
$\displaystyle \text{Answer:}$
$\displaystyle \frac{2}{x+1}+\frac{3}{2(x-2)}=\frac{23}{5x}$
$\displaystyle \Rightarrow 20x(x-2)+15x(x+1)=46(x+1)(x-2)$
$\displaystyle \Rightarrow 20x^2-40x+15x^2+15x=46(x^2-x-2)$
$\displaystyle \Rightarrow 35x^2-25x=46x^2-46x-92$
$\displaystyle \Rightarrow 11x^2-21x-92=0$
$\displaystyle \Rightarrow 11x^2-44x+23x-92=0$
$\displaystyle \Rightarrow 11x(x-4)+23(x-4)=0$
$\displaystyle \Rightarrow (x-4)(11x+23)=0$
$\displaystyle \Rightarrow x-4=0\text{ or }11x+23=0$
$\displaystyle \therefore x=4\text{ or }x=-\frac{23}{11}$
$\displaystyle \\$

$\displaystyle \textbf{Question 16: }\text{Solve the following quadratic equation by factorization:}$
$\displaystyle a^2x^2-3abx+2b^2=0$
$\displaystyle \text{Answer:}$
$\displaystyle a^2x^2-3abx+2b^2=0$
$\displaystyle \Rightarrow a^2x^2-abx-2abx+2b^2=0$
$\displaystyle \Rightarrow ax(ax-b)-2b(ax-b)=0$
$\displaystyle \Rightarrow (ax-b)(ax-2b)=0$
$\displaystyle \Rightarrow ax-b=0\text{ or }ax-2b=0$
$\displaystyle \therefore x=\frac{b}{a}\text{ or }x=\frac{2b}{a},\quad a\ne0$
$\displaystyle \\$

$\displaystyle \textbf{Question 17: }\text{Solve the following quadratic equation by factorization:}$
$\displaystyle 9x^2-6b^2x-(a^4-b^4)=0\hfill\text{[CBSE 2015]}$
$\displaystyle \text{Answer:}$
$\displaystyle 9x^2-6b^2x-(a^4-b^4)=0$
$\displaystyle \Rightarrow 9x^2-6b^2x-(a^2-b^2)(a^2+b^2)=0$
$\displaystyle \Rightarrow 9x^2-3(a^2+b^2)x+3(a^2-b^2)x$
$\displaystyle \qquad -(a^2-b^2)(a^2+b^2)=0$
$\displaystyle \Rightarrow 3x\{3x-(a^2+b^2)\}$
$\displaystyle \qquad +(a^2-b^2)\{3x-(a^2+b^2)\}=0$
$\displaystyle \Rightarrow \{3x-(a^2+b^2)\}\{3x+(a^2-b^2)\}=0$
$\displaystyle \Rightarrow 3x-(a^2+b^2)=0\text{ or }3x+(a^2-b^2)=0$
$\displaystyle \therefore x=\frac{a^2+b^2}{3}\text{ or }x=-\frac{a^2-b^2}{3}$
$\displaystyle \\$

$\displaystyle \textbf{Question 18: }\text{Solve the following quadratic equation by factorization:}$
$\displaystyle 4x^2+4bx-(a^2-b^2)=0$
$\displaystyle \text{Answer:}$
$\displaystyle 4x^2+4bx-(a^2-b^2)=0$
$\displaystyle \Rightarrow 4x^2+4bx-(a-b)(a+b)=0$
$\displaystyle \Rightarrow 4x^2-2(a-b)x+2(a+b)x-(a-b)(a+b)=0$
$\displaystyle \Rightarrow 2x\{2x-(a-b)\}+(a+b)\{2x-(a-b)\}=0$
$\displaystyle \Rightarrow \{2x-(a-b)\}\{2x+(a+b)\}=0$
$\displaystyle \Rightarrow 2x-(a-b)=0\text{ or }2x+(a+b)=0$
$\displaystyle \therefore x=\frac{a-b}{2}\text{ or }x=-\frac{a+b}{2}$
$\displaystyle \\$

$\displaystyle \textbf{Question 19: }\text{Solve the following quadratic equation by factorization:}$
$\displaystyle x^2+\left(a+\frac{1}{a}\right)x+1=0,\quad a\ne0$
$\displaystyle \text{Answer:}$
$\displaystyle x^2+\left(a+\frac{1}{a}\right)x+1=0$
$\displaystyle \Rightarrow x^2+ax+\frac{x}{a}+1=0$
$\displaystyle \Rightarrow x(x+a)+\frac{1}{a}(x+a)=0$
$\displaystyle \Rightarrow (x+a)\left(x+\frac{1}{a}\right)=0$
$\displaystyle \Rightarrow x+a=0\text{ or }x+\frac{1}{a}=0$
$\displaystyle \therefore x=-a\text{ or }x=-\frac{1}{a}$
$\displaystyle \\$

$\displaystyle \textbf{Question 20: }\text{Solve the following quadratic equation by factorization:}$
$\displaystyle abx^2+(b^2-ac)x-bc=0\hfill\text{[CBSE 2005]}$
$\displaystyle \text{Answer:}$
$\displaystyle abx^2+(b^2-ac)x-bc=0$
$\displaystyle \Rightarrow abx^2+b^2x-acx-bc=0$
$\displaystyle \Rightarrow bx(ax+b)-c(ax+b)=0$
$\displaystyle \Rightarrow (ax+b)(bx-c)=0$
$\displaystyle \Rightarrow ax+b=0\text{ or }bx-c=0$
$\displaystyle \therefore x=-\frac{b}{a}\text{ or }x=\frac{c}{b},\quad a\ne0,\ b\ne0$
$\displaystyle \\$

$\displaystyle \textbf{Question 21: }\text{Solve the following quadratic equation by factorization:}$
$\displaystyle 3\sqrt{5}x^2+25x-10\sqrt{5}=0\hfill\text{[CBSE 2015]}$
$\displaystyle \text{Answer:}$
$\displaystyle 3\sqrt{5}x^2+25x-10\sqrt{5}=0$
$\displaystyle \Rightarrow 3\sqrt{5}x^2+30x-5x-10\sqrt{5}=0$
$\displaystyle \Rightarrow 3\sqrt{5}x(x+2\sqrt{5})-5(x+2\sqrt{5})=0$
$\displaystyle \Rightarrow (x+2\sqrt{5})(3\sqrt{5}x-5)=0$
$\displaystyle \Rightarrow x+2\sqrt{5}=0\text{ or }3\sqrt{5}x-5=0$
$\displaystyle \therefore x=-2\sqrt{5}\text{ or }x=\frac{\sqrt{5}}{3}$
$\displaystyle \\$

$\displaystyle \textbf{Question 22: }\text{Solve the following quadratic equation by factorization:}$
$\displaystyle \sqrt{3}x^2-2\sqrt{2}x-2\sqrt{3}=0\hfill\text{[CBSE 2014]}$
$\displaystyle \text{Answer:}$
$\displaystyle \sqrt{3}x^2-2\sqrt{2}x-2\sqrt{3}=0$
$\displaystyle \Rightarrow \sqrt{3}x^2-3\sqrt{2}x+\sqrt{2}x-2\sqrt{3}=0$
$\displaystyle \Rightarrow \sqrt{3}x(x-\sqrt{6})+\sqrt{2}(x-\sqrt{6})=0$
$\displaystyle \Rightarrow (x-\sqrt{6})(\sqrt{3}x+\sqrt{2})=0$
$\displaystyle \Rightarrow x-\sqrt{6}=0\text{ or }\sqrt{3}x+\sqrt{2}=0$
$\displaystyle \therefore x=\sqrt{6}\text{ or }x=-\frac{\sqrt{2}}{\sqrt{3}}$
$\displaystyle \therefore x=\sqrt{6}\text{ or }x=-\frac{\sqrt{6}}{3}$
$\displaystyle \\$

$\displaystyle \textbf{Question 23: }\text{Solve the following quadratic equation by factorization:}$
$\displaystyle 3x^2-2\sqrt{6}x+2=0\hfill\text{CBSE 2010, 2012]}$
$\displaystyle \text{Answer:}$
$\displaystyle 3x^2-2\sqrt{6}x+2=0$
$\displaystyle \Rightarrow 3x^2-\sqrt{6}x-\sqrt{6}x+2=0$
$\displaystyle \Rightarrow \sqrt{3}x(\sqrt{3}x-\sqrt{2})-\sqrt{2}(\sqrt{3}x-\sqrt{2})=0$
$\displaystyle \Rightarrow (\sqrt{3}x-\sqrt{2})^2=0$
$\displaystyle \Rightarrow \sqrt{3}x-\sqrt{2}=0$
$\displaystyle \Rightarrow x=\frac{\sqrt{2}}{\sqrt{3}}=\frac{\sqrt{6}}{3}$
$\displaystyle \therefore \text{Both roots are equal and are }\frac{\sqrt{6}}{3}.$
$\displaystyle \\$

$\displaystyle \textbf{Question 24: }\text{Solve the following quadratic equation by factorization:}$
$\displaystyle \sqrt{2}x^2+7x+5\sqrt{2}=0\hfill\text{[CBSE 2017]}$
$\displaystyle \text{Answer:}$
$\displaystyle \sqrt{2}x^2+7x+5\sqrt{2}=0$
$\displaystyle \Rightarrow \sqrt{2}x^2+2x+5x+5\sqrt{2}=0$
$\displaystyle \Rightarrow \sqrt{2}x(x+\sqrt{2})+5(x+\sqrt{2})=0$
$\displaystyle \Rightarrow (x+\sqrt{2})(\sqrt{2}x+5)=0$
$\displaystyle \Rightarrow x+\sqrt{2}=0\text{ or }\sqrt{2}x+5=0$
$\displaystyle \therefore x=-\sqrt{2}\text{ or }x=-\frac{5}{\sqrt{2}}$
$\displaystyle \therefore x=-\sqrt{2}\text{ or }x=-\frac{5\sqrt{2}}{2}$
$\displaystyle \\$

$\displaystyle \textbf{Question 25: }\text{Solve the following quadratic equation by factorization:}$
$\displaystyle 3\left(\frac{7x+1}{5x-3}\right)-4\left(\frac{5x-3}{7x+1}\right)=11,\quad x\ne\frac{3}{5},-\frac{1}{7}$
$\displaystyle \hfill\text{[CBSE 2014]}$
$\displaystyle \text{Answer:}$
$\displaystyle 3\left(\frac{7x+1}{5x-3}\right)-4\left(\frac{5x-3}{7x+1}\right)=11$
$\displaystyle \Rightarrow 3(7x+1)^2-4(5x-3)^2=11(5x-3)(7x+1)$
$\displaystyle \Rightarrow 147x^2+42x+3-100x^2+120x-36$
$\displaystyle \qquad =385x^2-176x-33$
$\displaystyle \Rightarrow 47x^2+162x-33=385x^2-176x-33$
$\displaystyle \Rightarrow 338x^2-338x=0$
$\displaystyle \Rightarrow 338x(x-1)=0$
$\displaystyle \Rightarrow x=0\text{ or }x-1=0$
$\displaystyle \therefore x=0\text{ or }x=1$
$\displaystyle \\$

$\displaystyle \textbf{Question 26: }\text{Solve the following quadratic equation by factorization:}$
$\displaystyle \frac{x+1}{x-1}+\frac{x-2}{x+2}=4-\frac{2x+3}{x-2},\quad x\ne1,-2,2$
$\displaystyle \hfill\text{[CBSE 2016]}$
$\displaystyle \text{Answer:}$
$\displaystyle \frac{x+1}{x-1}+\frac{x-2}{x+2}=4-\frac{2x+3}{x-2}$
$\displaystyle \Rightarrow (x+1)(x+2)(x-2)+(x-2)^2(x-1)$
$\displaystyle \qquad =4(x-1)(x+2)(x-2)-(2x+3)(x-1)(x+2)$
$\displaystyle \Rightarrow 2x^3-3x^2-4x+6=2x^3-8x^2+15x+36$
$\displaystyle \Rightarrow 5x^2-19x-30=0$
$\displaystyle \Rightarrow 5x^2+25x-6x-30=0$
$\displaystyle \Rightarrow 5x(x+5)-6(x+5)=0$
$\displaystyle \Rightarrow (x+5)(5x-6)=0$
$\displaystyle \Rightarrow x+5=0\text{ or }5x-6=0$
$\displaystyle \therefore x=-5\text{ or }x=\frac{6}{5}$
$\displaystyle \\$

$\displaystyle \textbf{Question 27: }\text{Solve the following quadratic equation by factorization:}$
$\displaystyle \frac{a}{x-b}+\frac{b}{x-a}=2,\quad x\ne a,b\hfill\text{[CBSE 2016]}$
$\displaystyle \text{Answer:}$
$\displaystyle \frac{a}{x-b}+\frac{b}{x-a}=2$
$\displaystyle \Rightarrow a(x-a)+b(x-b)=2(x-a)(x-b)$
$\displaystyle \Rightarrow ax-a^2+bx-b^2=2x^2-2(a+b)x+2ab$
$\displaystyle \Rightarrow 2x^2-3(a+b)x+(a+b)^2=0$
$\displaystyle \Rightarrow 2x^2-2(a+b)x-(a+b)x+(a+b)^2=0$
$\displaystyle \Rightarrow 2x\{x-(a+b)\}-(a+b)\{x-(a+b)\}=0$
$\displaystyle \Rightarrow \{x-(a+b)\}\{2x-(a+b)\}=0$
$\displaystyle \Rightarrow x-(a+b)=0\text{ or }2x-(a+b)=0$
$\displaystyle \therefore x=a+b\text{ or }x=\frac{a+b}{2}$
$\displaystyle \\$

$\displaystyle \textbf{Question 28: }\text{Solve the following quadratic equation by factorization:}$
$\displaystyle \frac{a}{x-a}+\frac{b}{x-b}=\frac{2c}{x-c},\quad x\ne a,b,c\hfill\text{[CBSE 2016]}$
$\displaystyle \text{Answer:}$
$\displaystyle \frac{a}{x-a}+\frac{b}{x-b}=\frac{2c}{x-c}$
$\displaystyle \Rightarrow a(x-b)(x-c)+b(x-a)(x-c)=2c(x-a)(x-b)$
$\displaystyle \Rightarrow (a+b)x^2-(a+b)cx-2abx+2abc$
$\displaystyle \qquad =2cx^2-2c(a+b)x+2abc$
$\displaystyle \Rightarrow (a+b-2c)x^2+\{c(a+b)-2ab\}x=0$
$\displaystyle \Rightarrow x\left[\{a+b-2c\}x+c(a+b)-2ab\right]=0$
$\displaystyle \Rightarrow x=0\text{ or }(a+b-2c)x=2ab-c(a+b)$
$\displaystyle \therefore x=0\text{ or }x=\frac{2ab-c(a+b)}{a+b-2c},\quad a+b\ne2c$
$\displaystyle \\$

$\displaystyle \textbf{Question 29: }\text{Solve the following quadratic equation by factorization:}$
$\displaystyle \frac{1}{2a+b+2x}=\frac{1}{2a}+\frac{1}{b}+\frac{1}{2x}\hfill\text{[CBSE 2013]}$
$\displaystyle \text{Answer:}$
$\displaystyle \frac{1}{2a+b+2x}=\frac{1}{2a}+\frac{1}{b}+\frac{1}{2x}$
$\displaystyle \Rightarrow \frac{1}{2a+b+2x}=\frac{bx+2ax+ab}{2abx}$
$\displaystyle \Rightarrow 2abx=(2a+b+2x)(bx+2ax+ab)$
$\displaystyle \Rightarrow 2abx=(2a+b+2x)\{(2a+b)x+ab\}$
$\displaystyle \Rightarrow 2(2a+b)x^2+(2a+b)^2x+ab(2a+b)=0$
$\displaystyle \Rightarrow (2a+b)\{2x^2+(2a+b)x+ab\}=0$
$\displaystyle \Rightarrow 2x^2+(2a+b)x+ab=0$
$\displaystyle \Rightarrow 2x^2+2ax+bx+ab=0$
$\displaystyle \Rightarrow 2x(x+a)+b(x+a)=0$
$\displaystyle \Rightarrow (x+a)(2x+b)=0$
$\displaystyle \Rightarrow x+a=0\text{ or }2x+b=0$
$\displaystyle \therefore x=-a\text{ or }x=-\frac{b}{2}$
$\displaystyle \\$

$\displaystyle \textbf{Question 30: }\text{Solve the following quadratic equation by factorization:}$
$\displaystyle (x-5)(x-6)=\frac{25}{(24)^2}$
$\displaystyle \text{Answer:}$
$\displaystyle (x-5)(x-6)=\frac{25}{576}$
$\displaystyle \Rightarrow x^2-11x+30-\frac{25}{576}=0$
$\displaystyle \Rightarrow \left(x-\frac{145}{24}\right)\left(x-\frac{119}{24}\right)=0$
$\displaystyle \Rightarrow x-\frac{145}{24}=0\text{ or }x-\frac{119}{24}=0$
$\displaystyle \therefore x=\frac{145}{24}\text{ or }x=\frac{119}{24}$
$\displaystyle \\$

$\displaystyle \textbf{Question 31: }\text{Solve the following quadratic equation by factorization:}$
$\displaystyle 7x+\frac{3}{x}=35\frac{3}{5},\quad x\ne0$
$\displaystyle \text{Answer:}$
$\displaystyle 7x+\frac{3}{x}=35\frac{3}{5}$
$\displaystyle \Rightarrow 7x+\frac{3}{x}=\frac{178}{5}$
$\displaystyle \Rightarrow 35x^2+15=178x$
$\displaystyle \Rightarrow 35x^2-178x+15=0$
$\displaystyle \Rightarrow 35x^2-175x-3x+15=0$
$\displaystyle \Rightarrow 35x(x-5)-3(x-5)=0$
$\displaystyle \Rightarrow (x-5)(35x-3)=0$
$\displaystyle \Rightarrow x-5=0\text{ or }35x-3=0$
$\displaystyle \therefore x=5\text{ or }x=\frac{3}{35}$
$\displaystyle \\$