JEE Main 2016MathematicsQuadratic EquationEasyMCQ

JEE Main 2016 — Quadratic Equation Question with Solution

JEE Main 2016 (09 Apr Online)

Question

If the equations x2+bx-1=0 and x2+x+b=0 have a common root different from -1, then b is equal to :

Choose an option

Show full solutionCorrect option: C
Correct answer
C3

Step-by-step explanation

Let common root be x
x2+bx-1=0     ...1

x2+x+b=0       ...2

1-2, we get x=b+1b-1

Put x in Equation 1, we get

b+1b-12+b+1b-1+b=0

⇒ b+12 + b+1 b-1+bb-12=0

⇒ b2+1+2b+b2-1+bb2-2b+1=0

⇒ 2b2+2b+b3-2b2+b=0

⇒  b3+3b=0
⇒   bb2+3=0⇒b=0 or b2=-3

When  b=0 then common root x= -1
But the given common root x≠-1

∴ b2= -3

b= ± 3i

b= 3

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 2016, 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.