Question

Solve the following quadratic equations by factorization method :
$\frac{1}{a+b+x}=\frac{1}{a}+\frac{1}{b}+\frac{1}{x}$

Answer 1

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

$⟹\frac{1}{a+b+x}=\frac{bx+ax+ab}{abx}$

$⟹abx=abx+a^{2}x+a^{2}b+b^{2}x+abx+a{b}^2+b{x}^2+a{x}^2+abx$

$⟹x^2(a+b)+x(2ab+a^2+b^2)+(a^{2}b+a{b}^2)=0$

$⟹x^2(a+b)+x(a+b{)}^2+ab(a+b)=0$

$⟹x^2+x(a+b)+ab=0$

$⟹x^2+ax+bx+ab=0$

$⟹(x+a)(x+b)=0$
$\therefore x=-a$, $x=-b$