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Q.
The roots of the equation $ |{{x}^{2}}-x-6|=x+2 $ are
Bihar CECEBihar CECE 2004
Solution:
We have $\left|x^{2}-x-6\right|=x+2 \,\,\,...$(i)
or $|(x-3)(x+2)|=x+2$
Case I: When $-2 \leq x$ or $x \geq 3$
From Eq. (i) $x^{2}-x-6=x+2 $
$\Rightarrow x^{2}-2 x-8=0$
$ \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 $-2 \leq x$ or $x \geq 3$
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 $\{-2,2,4\}$.
