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JEE Main 202124 FebruaryMorning Shift
Exam
JEE Main
Year
2021
Date
24 February
Shift
Morning Shift
Subject
Mathematics
Chapter
Quadratic Equation

JEE Main 2021 · 24 February · Morning Shift · Mathematics · Quadratic Equation

Let p and q be two positive numbers such that p + q = 2 and p ^ 4 +...

Let pp and qq be two positive numbers such that p+q=2p + q = 2 and p4+q4=272{p}^{4} + {q}^{4} = 272. Then pp and qq are roots of the equation:

Options

  1. Ax2−2x+2=0{x}^{2} - 2x + 2 = 0
  2. Bx2−2x+8=0{x}^{2} - 2x + 8 = 0
  3. Cx2−2x+136=0{x}^{2} - 2x + 136 = 0
  4. Dx2−2x+16=0{x}^{2} - 2x + 16 = 0

Correct answer

D. x2−2x+16=0{x}^{2} - 2x + 16 = 0

Step-by-step solution

Consider (p2+q2)2−2p2q2=272{\left({p}^{2} + {q}^{2}\right)}^{2} - 2{p}^{2}{q}^{2} = 272

((p+q)2−2pq)2−2p2q2=272{\left({\left(p + q\right)}^{2} - 2pq\right)}^{2} - 2{p}^{2}{q}^{2} = 272

16−16pq+2p2q2=27216 - 16pq + 2{p}^{2}{q}^{2} = 272

(pq)2−8pq−128=0{\left(pq\right)}^{2} - 8pq - 128 = 0

(pq)2−8pq−128=0{\left(pq\right)}^{2} - 8pq - 128 = 0

pq=8±242=16,−8pq = \dfrac{8 \pm 24}{2} = 16, - 8

∴pq=16∴ pq = 16

∴∴Required equation

:x2−(2)x+16=0: {x}^{2} - \left(2\right)x + 16 = 0

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