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Mathematics · Quadratic Equations

JEE Main 2025 — 28 January, Morning Shift — Question 15

The sum, of the squares of all the roots of the equation x2+∣2x−3∣−4=0x^{2}+|2 x-3|-4=0, is :

  1. Option A:

    3(3−2)3(3-\sqrt{2})

  2. Option B:

    6(3−2)6(3-\sqrt{2})

  3. Option C:

    6(2−2)6(2-\sqrt{2})

    Correct
  4. Option D:

    3(2−2)3(2-\sqrt{2})

Answer: C

Step-by-step solution

x2+∣2x−3∣−4=0x^{2}+|2 x-3|-4=0

Case I: x≥32\mathrm{x} \geq \frac{3}{2}

x2+2x−3−4=0x^{2}+2 x-3-4=0 x=22−1x=2 \sqrt{2}-1

Case II : x<32\mathrm{x}<\frac{3}{2}

x2+3−2x−4=0x2−2x−1=0x=1−2\begin{aligned} & x^{2}+3-2 x-4=0 \\& x^{2}-2 x-1=0 \\& x=1-\sqrt{2} \end{aligned}

Sum of squares =(22−1)2+(1−2)2=(2 \sqrt{2}-1)^{2}+(1-\sqrt{2})^{2}

=8−42+1+1−22+2=6(2−2)\begin{aligned} & =8-4 \sqrt{2}+1+1-2 \sqrt{2}+2 \\& =6(2-\sqrt{2}) \end{aligned}

Answer key and solution verified before publishing.

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Exam
JEE Main 2025
Paper
28 January, Morning Shift
Subject
Mathematics
Chapter
Quadratic Equations
Topic
Nature of Roots of Quadratic Equation
The sum, of the squares of all the roots of the equation x 2 + 2 x-3… | JEE Main 2025 PYQ with Solution · DhiX AI