JEE Main 2022MathematicsQuadratic EquationMediumNumerical

JEE Main 2022 — Quadratic Equation Question with Solution

JEE Main 2022 (26 Jul Shift 1)

Question

If for some p,q,r∈R, all have positive sign, one of the roots of the equation p2+q2x2-2qp+rx+q2+r2=0 is also a root of the equation x2+2x-8=0, then q2+r2p2 is equal to-

Enter your answer

Show full solutionCorrect answer: 272
Correct answer
272

Step-by-step explanation

Given, p2+q2x2-2qp+rx+q2+r2=0

On simplifying we get, px-q2+qx-r2=0

⇒px-q=0 & qx-r=0

⇒x=qp=rq

⇒  x=qp=rq=4    [because roots of equation x2-2x-8=0 are 4 or -2]

As p,q,r are positive, so x must be 4.

Now, q=4p and r=4q=16p

So, q2+r2p2=4p2+4×4p2p2=16+256=272.

Practice this on the real CBT interface

Solve this JEE Main question (and the rest of the Quadratic Equation chapter) on PrepSharp's TCS iON-style CBT player — with timer, bookmarks and session analytics.

Solve interactively →

About this question

This is a previous-year question from JEE Main 2022, covering the Quadratic Equation chapter of Mathematics. PrepSharp catalogues every PYQ from JEE Main with a verified answer key and step-by-step solution prepared by IIT alumni — so you can search by chapter, topic or year and revise efficiently.