Question

Given that $\frac{x^2-5x+6}{x^2+6x-16}=\frac{x^2-2x-3}{-6}$, then all possible values of $x$ are:

• A. $-7$
• B. $-5$
• C. $0$
• D. $6$

Answer 1

The correct option is A $-7$

Given, $\frac{x^2-5x+6}{x^2+6x-16}=\frac{x^2-2x-3}{-6}$

$\Rightarrow \frac{x^2-3x-2x+6}{x^2+8x-2x-16}=\frac{x^2-3x+x-3}{-6}$

$\Rightarrow \frac{\left(x\right(x-3)-2(x-3\left)\right)}{x(x+8)-2(x+8)}=\frac{x(x-3)+1(x-3)}{-6}$

$\Rightarrow \frac{(x-3)(x-2)}{(x+8)(x-2)}=\frac{(x-3)(x+1)}{-6}$

$\Rightarrow \frac{1}{x+8}=\frac{x+1}{-6}$

$\Rightarrow (x+1)(x+8)=-6$

$\Rightarrow x^2+x+8x+8=-6$

$\Rightarrow x^2+9x+14=0$

$\Rightarrow x^2+7x+2x+14=0$

$\Rightarrow x(x+7)+2(x+7)=0$

$\Rightarrow (x+7)(x+2)=0$

Then $x+7=0$ or $x=-7$

And $x+2=0$ or $x=-2$