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

Answer 1

(b) 1

$$\begin{gathered} \frac{(x+2)(x-5)}{(x-3)(x+6)}=\frac{(x-2)}{(x+4)} \\ \Rightarrow (x^2-3x-10)(x+4)=(x^2+3x-18)(x-2) \\ \Rightarrow x^3+4x^2-3x^2-12x-10x-40=x^3-2x^2+3x^2-6x-18x+36 \\ \Rightarrow x^2-22x-40=x^2-24x+36 \\ \Rightarrow 2x=76 \\ \Rightarrow x=38 \end{gathered}$$
Hence, the equation has only 1 root.