Explanation:
B \(\left|x^2-x-6\right|=x+2\)
Case- I \(\quad x^2-x-6\lt 0\)
\(\Rightarrow(\mathrm{x}-3)(\mathrm{x}+2)\lt 0\)
\(\Rightarrow-2\lt x\lt 3\)
In this case, the equation becomes
\(\mathrm{x}^2-\mathrm{x}-6=-(\mathrm{x}+2)\)
Or \(\mathrm{x}^2-4=0\)
\(\therefore \mathrm{x}= \pm 2\)
Clearly, \(x=2\) satisfies the domain of the equation in
this case, so \(x=2\), is a solution.
Case- II
\(\mathrm{x}^2-\mathrm{x}-6 \geq 0\)
So,
\(\mathrm{x} \leq-2 \text { or } \mathrm{x} \geq 3\)
In this case, the equation becomes
\(x^2-x-6=0=x+2\)
\(x^2-2 x-8=0\)
\(=x^2-4 x+2 x-8=0\)
\(=x(x-4)+2(x-4)=0\)
\(=(x+2)(x-4)=0\)
\(\Rightarrow x=-2,4\)
Both these values lie in the domain of the equation in this case, So \(\mathrm{x}=-2,4\) are the roots.
Hence, roots are \(\mathrm{x}=-2,2,4\)