JEE Main 2020MathematicsQuadratic EquationEasyMCQ

JEE Main 2020 — Quadratic Equation Question with Solution

JEE Main 2020 (05 Sep Shift 2)

Question

If  α and β are the roots of the equation, 7x2-3x-2=0, then the value ofα1-α2+β1-β2 is equal to:

Choose an option

Show full solutionCorrect option: D
Correct answer
D2716

Step-by-step explanation

Given α and β are the roots of the equation,7x2-3x-2=0, then 

α+β=37 ; αβ=-27

Now 

α1-α2+β1-β2=(α+β)-αβ(α+β)1-α21-β2=(α+β)-αβ(α+β)1+(αβ)2-α2+β2

⇒(α+β)-αβ(α+β)1+(αβ)2-(α+β)2+2αβ=37+27371+272-372-227=2716

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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.