Question

Find the roots of the equation $\frac{1}{a+b+x}-\frac{1}{x}=\frac{1}{a}+\frac{a}{b}$.

• A. $a,-b$
• B. $-a,b$
• C. $a,b$
• D. $-a,-b$

Answer 1

The correct option is D $-a,-b$

Given equation is: $\frac{1}{a+b+x}-\frac{1}{x}=\frac{1}{a}+\frac{a}{b}$

Simplifying, we get
$\frac{x-(a+b+x)}{x(a+b+x)}=\frac{b+a}{ab}$
$\frac{-(a+b)}{(a+b)x+x^2}=\frac{b+a}{ab}$
$-(a+b)ab=(a+b{)}^{2}x+(a+b)x^2$
$(a+b)x^2+(a+b{)}^{2}x+(a+b)ab=0$
$(a+b)[x^2+(a+b)x+ab]=0$
$(a+b)\left[\right(x+a\left)\right(x+b\left)\right]=0$
Hence
$x=-a,-b$ and $a=-b$.