The number of roots of the equation
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.

 

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