Question:
The number of roots of the equation $\frac{(x+2)(x-5)}{(x-3)(x+6)}=\frac{x-2}{x+4}$ is
(a) 0
(b) 1
(c) 2
(d) 3
Solution:
(b) 1
$\frac{(x+2)(x-5)}{(x-3)(x+6)}=\frac{(x-2)}{(x+4)}$
$\Rightarrow\left(x^{2}-3 x-10\right)(x+4)=\left(x^{2}+3 x-18\right)(x-2)$
$\Rightarrow x^{3}+4 x^{2}-3 x^{2}-12 x-10 x-40=x^{3}-2 x^{2}+3 x^{2}-6 x-18 x+36$
$\Rightarrow x^{2}-22 x-40=x^{2}-24 x+36$
$\Rightarrow 2 x=76$
$\Rightarrow x=38$
Hence, the equation has only 1 root.