Question

The roots of the equation $|x^2-x-6|=x+2$ are

• A. -2,1,4
• B. 0,2,4
• C. 0,1,4
• D. -2,2,4

Answer 1

Answer: (D)

We have $\left|x^2-x-6\right|=x+2...$(i)
or $|(x-3)(x+2)|=x+2$
Case I: When $-2\le x$ or $x\ge 3$
From Eq. (i) $x^2-x-6=x+2$
$\Rightarrow x^2-2x-8=0$
$\Rightarrow (x-4)(x+2)=0$
$\Rightarrow x=-2,4$
Case II : When $-2<x<3$
From Eq. (i) $-\left(x^2-x-6\right)=x+2$
$\Rightarrow -x^2+x+6=x+2$
$\Rightarrow -x^2=-4$
$\Rightarrow x=\pm 2$
$\Rightarrow x=2(\because x=-2$ not lies in the interval)
Hence, the roots are $\{-2,2,4\}$ Alternate Solution :
Case I : When $-2\le x$ or $x\ge 3$
Let $y=x^2-x-6=x+2$
Now, $y=x^2-x-6$
$\left(x^2-x+\frac{1}{4}\right)=y+6+\frac{1}{4}$

${\left(x-\frac{1}{2}\right)}^2=y+\frac{25}{4}$
and $y=x+2$
$\Rightarrow \frac{y}{2}-\frac{x}{2}=1$
Case II : When $-2<x<3$
Let $y=-\left(x^2-x-6\right)=x+2$
Now, $y=-\left(x^2-x-6\right)$

$-\left(x^2-x+\frac{1}{4}\right)=y-6-\frac{1}{4}$
${\left(x-\frac{1}{2}\right)}^2=-\left(y-\frac{25}{4}\right)$
and $y=x+2$
It is clear from the above two figures that the curve will intersect the line in three different noints ie $\{-2,2,4\}$.