Question

If | $x^2-x-6|=x+2$, then the values of $x$ are

• A. -2, 2, -4
• B. -2, 2, 4
• C. 3, 2, -2
• D. 4, 4, 3

Answer 1

Answer: (B)

$\left|x^2-x-6\right|=x+2$, then
Case I $x^2-x-6<0$
$\Rightarrow (x-3)(x+2)<0$
$\Rightarrow -2<x<3$
In this case, the equation becomes
$x^2-x-6=-x-2$
or $x^2-4=0$
$\therefore x=\pm 2$
Clearly, $x=2$ satisfies the domain of the equation in this case.
So, $x=2$ is a solution.
Case II $x^2-x-6\ge 0$
So, $x\le -2$ or $x\ge 3$
Then, equation reduces to $x^2-x-6=0=x+2$
i.e., $x^2-2x-8=0$ or $x=-2,4$
Both these values lies in the domain of the equation in this case,
so $x=-2,4$ are the roots.
Hence, roots are $x=-2,2,4$.