Question

Solve the following quadratic equations by factorization method: [3 MARKS]

$\frac{1}{a+b+x}=\frac{1}{a}+\frac{1}{b}+\frac{1}{x},a+b\ne 0$

Answer 1

Concept : 1 Mark
Application : 2 Marks

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

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

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

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

$\Rightarrow -ab=ax+bx+x^2$

$\Rightarrow x^2+(a+b)x+ab=0$ $[\because a+b\ne 0]$

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

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

$\Rightarrow (x+a)(x+b)=0\Rightarrow x+a=0or,x+b=0\Rightarrow x=-aor,x=-b$