The sum of the squares of the roots of |x+2|^2+|x-2|-2=0 and the squares of the roots of x^2-2|x-3|-5=0 , is

Correct answer: B. 36 aligned & |x-2|^2+2|x-2|-|x-2|-2=0 & (|x-2|+2)(|x-2|-1)=0 & |x-2|=1 & x=2 ± 1=3,1 & sum of square of roots =9+1=10 & x^2-2|x-3|-5=0 & Case-I x-3 ≥ 0 & x^2-2 x+1=0 & (x-1)^2=0 & x=1 aligned But x ≥ 3 x ∈ φ Case-II x-3 0 Real \& distinct roots f(x)=x^2+2 x-11 f(3) > 0, -p 2 a =-1 < 3 both roots < 3 , both roots acceptableSum of square of roots =(α+β)^2-2 α β aligned & =4+22=26 & Final sum =10+26=36 aligned

JEE Main 20258 AprilEvening Shift

Exam: JEE Main
Year: 2025
Date: 8 April
Shift: Evening Shift
Subject: Mathematics
Chapter: Quadratic Equation

JEE Main 2025 · 8 April · Evening Shift · Mathematics · Quadratic Equation

The sum of the squares of the roots of |x+2|^2+|x-2|-2=0 and the...

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

Options

Correct answer

B. 36

Step-by-step solution

\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 < 0 x^2+2 x-11=0, D > 0 \Rightarrow

Real \& distinct roots

f(x)=x^2+2 x-11 \mathrm{f}(3) > 0, \frac{-\mathrm{p}}{2 \mathrm{a}}=-1 < 3 \Rightarrow

both roots

< 3

, both roots acceptableSum of square of roots

=(\alpha+\beta)^2-2 \alpha \beta

\begin{aligned}& =4+22=26 \\ & \Rightarrow \text{ Final sum }=10+26=36\end{aligned}

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