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. 37715

Answer 1

Answer: (B)

-2, 2, 4

Answer 2

$\left|x^{2}-x-6\right|=x+2$, 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 $x^{2}-x-6 \geq 0$ So, $x \leq-2$ or $x \geq 3$ Then, equation reduces to $x^{2}-x-6=0=x+2$ i.e., $x^{2}-2 x-8=0$ 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 $x=-2,2,4$.