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Quadratic Equation and Inequalities question

2025 · 28 Jan · Shift 1 · Q28
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  5. /2025 · 28 Jan · Shift 1 · Q28

Quadratic Equation and Inequalities question

2025 · 28 Jan · Shift 1 · Q28

JEE MainMathematicsQuadratic Equation and InequalitiesMCQ+4 / −1
The sum, of the squares of all the roots of the equation x2+∣2x−3∣−4=0x^2+|2 x-3|-4=0x2+∣2x−3∣−4=0, is
  1. A
    6(2−2)6(2-\sqrt{2})6(2−2​)
  2. B
    3(3−2)3(3-\sqrt{2})3(3−2​)
  3. C
    3(2−2)3(2-\sqrt{2})3(2−2​)
  4. D
    6(3−2)6(3-\sqrt{2})6(3−2​)
View written solutionFree

Correct answer: A

  1. We need to solve x2+∣2x−3∣−4=0x^2+|2x-3|-4=0x2+∣2x−3∣−4=0 and then find the sum of the squares of all its roots.

  2. Split by the absolute value:

  • If 2x−3≥02x-3\ge 02x−3≥0, i.e. x≥32x\ge \frac{3}{2}x≥23​, then ∣2x−3∣=2x−3.|2x-3|=2x-3.∣2x−3∣=2x−3. The equation becomes x2+(2x−3)−4=0x^2+(2x-3)-4=0x2+(2x−3)−4=0 x2+2x−7=0.x^2+2x-7=0.x2+2x−7=0. Solving, x=−2±4+282=−2±322=−1±22.x=\frac{-2\pm\sqrt{4+28}}{2}=\frac{-2\pm\sqrt{32}}{2}=-1\pm 2\sqrt{2}.x=2−2±4+28​​=2−2±32​​=−1±22​. Now apply the condition x≥32x\ge \frac{3}{2}x≥23​:

    • −1+22-1+2\sqrt{2}−1+22​ is valid since −1+22≈1.828>1.5-1+2\sqrt{2}\approx 1.828>1.5−1+22​≈1.828>1.5.
    • −1−22-1-2\sqrt{2}−1−22​ is invalid.

    So one root is x1=−1+22.x_1=-1+2\sqrt{2}.x1​=−1+22​.

  1. If 2x−3<02x-3<02x−3<0, i.e. x<32x<\frac{3}{2}x<23​, then ∣2x−3∣=−(2x−3)=−2x+3.|2x-3|=-(2x-3)=-2x+3.∣2x−3∣=−(2x−3)=−2x+3. The equation becomes x2+(−2x+3)−4=0x^2+(-2x+3)-4=0x2+(−2x+3)−4=0 x2−2x−1=0.x^2-2x-1=0.x2−2x−1=0. Solving, x=2±4+42=2±82=1±2.x=\frac{2\pm\sqrt{4+4}}{2}=\frac{2\pm\sqrt{8}}{2}=1\pm\sqrt{2}.x=22±4+4​​=22±8​​=1±2​. Now apply the condition x<32x<\frac{3}{2}x<23​:
  • 1+2≈2.4141+\sqrt{2}\approx 2.4141+2​≈2.414 is invalid.
  • 1−21-\sqrt{2}1−2​ is valid.

So the second root is x2=1−2.x_2=1-\sqrt{2}.x2​=1−2​.

  1. Now compute the sum of squares: x12+x22=(−1+22)2+(1−2)2.x_1^2+x_2^2 = (-1+2\sqrt{2})^2+(1-\sqrt{2})^2.x12​+x22​=(−1+22​)2+(1−2​)2.

First, (−1+22)2=1+8−42=9−42.(-1+2\sqrt{2})^2=1+8-4\sqrt{2}=9-4\sqrt{2}.(−1+22​)2=1+8−42​=9−42​.

Second, (1−2)2=1+2−22=3−22.(1-\sqrt{2})^2=1+2-2\sqrt{2}=3-2\sqrt{2}.(1−2​)2=1+2−22​=3−22​.

Therefore, x12+x22=(9−42)+(3−22)=12−62.x_1^2+x_2^2=(9-4\sqrt{2})+(3-2\sqrt{2})=12-6\sqrt{2}.x12​+x22​=(9−42​)+(3−22​)=12−62​. Factorizing, 12−62=6(2−2).12-6\sqrt{2}=6(2-\sqrt{2}).12−62​=6(2−2​).

  1. Hence the required sum is 6(2−2).\boxed{6(2-\sqrt{2})}.6(2−2​)​. So the correct option is A.
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