Question

The equation $\frac{(x+2)(x-5)}{(x-3)(x+6)}=\frac{x-2}{x+4}$ has

• A. $3$ roots
• B. $2$ roots
• C. $1$ root
• D. no root

Answer 1

Answer: (C)

$1$ root

The given equation is

$\frac{(x+2)(x-5)}{(x-3)(x+6)}=\frac{(x-2)}{(x+4)}$

$\Rightarrow (x+4)(x+2)(x-5)=(x-2)(x-3)(x+6)$

$\Rightarrow (x+4)(x+2)(x-5)-(x-2)(x-3)(x+6)=0$

$\Rightarrow (x^3+x^2-22x-40)-(x^3+x^2-24x+36)=0$

$\Rightarrow 2x-76=0\Rightarrow \frac{76}{2}$

i.e. $x=38,\therefore$ the given equation has only one root.