JEE Main 2019MathematicsQuadratic EquationMediumMCQ

JEE Main 2019 — Quadratic Equation Question with Solution

JEE Main 2019 (09 Jan Shift 1)

Question

Let α and β be the roots of the equation x2+2x+2=0, then α15+β15 is equal to

Choose an option

Show full solutionCorrect option: D
Correct answer
D-256

Step-by-step explanation

Let the two roots of  x2+2x+2=0 be α and β

⇒α,β=-2±4-82=-1±i

⇒α,β=2-12±i12

=2cos3π4±isin3π4

⇒α15+β15=2cos3π4+isin3π415+2cos 3π4-isin3π415

       ⇒α15+β15=2152cos45π4+isin45π4+cos45π4-isin45π4

⇒α15+β15=2.2152.cos45π4=2172.cos5π4=-2172.12

⇒α15+β15=-256

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About this question

This is a previous-year question from JEE Main 2019, 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.