Question

The number of solutions of equation $\mid x^2-x-6\mid =x+2$, where $x$ is a real number is

• A. $1$
• B. $2$
• C. $3$
• D. $4$

Answer 1

The correct option is C $3$

Here $\mid x^2-x-6\mid$ occurs and $x^2-x-6=0$
$\Rightarrow x=-2,3$
Given equation $\mid x^2-x-6\mid =x+2$ .....(1)
Case 1. When $x<-2$ or x > 3. In this case $x^2-x-6>0$
$\therefore \mid x^2-x-6\mid =x^2-x-6$
Now, equation (1) becomes $x^2-x-6=x+2\Rightarrow x^2-2x-8=0\therefore x=-2,4$
Only $x=4$ is acceptable $(\because x<-2$ or $x>3)$
Case 2. When $-2<x<3.x^2-x-6<0$
$\therefore \mid x^2-x-6\mid =-(x^2-x-6)$
Now, equation (1) becomes
$-(x^2-x-6)=x+2$
$\Rightarrow x^2-4=0\Rightarrow x=+2,-2$
Since, $x=-2$ is not acceptable in this case.
Case 3. Clearly, $x=-2$ satisfies equation (1)
But $x=3$ does not satisfy it
Thus, $x=-2,2,4$.
Note: Equality sign may be taken in case 1 and 2 and in that case, case 3 will not be required.