The sum of the squares of the roots of $|\mathrm{x}+2|^2+|\mathrm{x}-2|-2=0$ and the squares of the roots of $x^2-2|x-3|-5=0$, is

Question

Aicharya · Accepted Answer

$\begin{aligned} & |\mathrm{x}-2|^2+2|\mathrm{x}-2|-|\mathrm{x}-2|-2=0 \\ & \Rightarrow(|\mathrm{x}-2|+2)(|\mathrm{x}-2|-1)=0 \\ & \Rightarrow|\mathrm{x}-2|=1 \\ & \Rightarrow \mathrm{x}=2 \pm 1=3,1 \\ & \Rightarrow \text { sum of square of roots }=9+1=10 \\ & \mathrm{x}^2-2|\mathrm{x}-3|-5=0 \\ & \text { Case-I } \mathrm{x}-3 \geq 0 \\ & \Rightarrow \mathrm{x}^2-2 \mathrm{x}+1=0 \\ & \Rightarrow(\mathrm{x}-1)^2=0 \\ & \Rightarrow \mathrm{x}=1\end{aligned}$
But $x \geq 3$
$\Rightarrow \mathrm{x} \in \phi$
Case-II $\mathrm{x}-3 \lt 0$
$x^2+2 x-11=0, D \gt 0 \Rightarrow$ Real \& distinct roots
$f(x)=x^2+2 x-11$
$\mathrm{f}(3) \gt 0, \frac{-\mathrm{p}}{2 \mathrm{a}}=-1 \lt 3$
$\Rightarrow$ both roots $ \lt 3$, both roots acceptable
Sum of square of roots $=(\alpha+\beta)^2-2 \alpha \beta$
$\begin{aligned}
& =4+22=26 \\ & \Rightarrow \text { Final sum }=10+26=36
\end{aligned}$ ~

The sum of the squares of the roots of $|\mathrm{x}+2|^2+|\mathrm{x}-2|-2=0$ and the squares of the roots of…

Solution