Question

If $a,b,(a>b)$ are the solution of the equation $|x{|}^2-|x|+4=2x^2-3|x|+1$, then value of $a-b$ is

Answer 1

$|x{|}^2-|x|+4=2x^2-3|x|+1$
$\Rightarrow |x{|}^2-|x|+4=2|x{|}^2-3|x|+1(\because x^2=|x{|}^2)$
$\Rightarrow |x{|}^2-2|x|-3=0$
Let $|x|=t$
Now, $t^2-2t-3=0$
$\Rightarrow (t-3)(t+1)=0$
$\Rightarrow t=-1,3$
$\Rightarrow |x|=-1,$ (not possible)
and
$|x|=3$
$\Rightarrow x=-3,+3$
$\Rightarrow a=3,b=-3$
$\Rightarrow a+b=3-(-3)=6$