JEE Main 2020MathematicsQuadratic EquationMediumMCQ

JEE Main 2020 — Quadratic Equation Question with Solution

JEE Main 2020 (09 Jan Shift 2)

Question

Let a,b∈R,a≠0 be such that the equation, ax2-2bx+5=0 has a repeated root α, which is also a root of the equation, x2-2bx-10=0. If β is the other root of this equation, then α2+β2 is equal to:

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Show full solutionCorrect option: A
Correct answer
A25

Step-by-step explanation

Given, ax2-2bx+5=0 has repeated root α.

∴2α=2ba⇒α=baand α2=5a⇒b2a2=5a

⇒b2=5a ...i a≠0

α+β=2b ...ii

and αβ=-10  ...iii

α=ba is also root of x2-2bx-10=0

⇒b2-2ab2-10a2=0

by i⇒5a-10a2-10a2=0

⇒20a2=5a

⇒a=14 and b2=54

Now α2+β2=α+β2-2αβ

=5+20

=25

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About this question

This is a previous-year question from JEE Main 2020, 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.