Question

Solve the given quadratic equation by factorization method
$\frac{1}{a+b+x}=\frac{1}{x}+\frac{1}{a}+\frac{1}{b},a+b\ne 0$

Answer 1

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

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

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

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

$\Rightarrow x(a+b+x)+ab=0\Rightarrow x^2+ax+bx+ab=0\Rightarrow x(x+a)+b(x+a)=0\Rightarrow (x+a)(x+b)=0$

$\Rightarrow x+a=0orx+b=0\Rightarrow x=-aorx=-b$