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JEE Main 20169 AprilOnline
Exam
JEE Main
Year
2016
Date
9 April
Mode
Online
Subject
Mathematics
Chapter
Quadratic Equation

JEE Main 2016 · 9 April · Online · Mathematics · Quadratic Equation

If the equations x ^ 2 + bx - 1 = 0 and x ^ 2 + x + b = 0 have a...

If the equations x2+bx−1=0andx2+x+b=0{x}^{2} + bx - 1 = 0 and {x}^{2} + x + b = 0 have a common root different from −1,then∣b∣- 1, then \left|b\right| is equal to :

Options

  1. A22
  2. B33
  3. C3\sqrt{3}
  4. D2\sqrt{2}

Correct answer

C. 3\sqrt{3}

Step-by-step solution

Let common root be xx
x2+bx−1=0{x}^{2} + bx - 1 = 0 ...\left(1\right)<br/><br/><br /><br />{x}^{2} + x + b = 0...(2)...\left(2\right)

\left(1\right) - \left(2\right),weget, we getx = \dfrac{b + 1}{b - 1}<br/><br/>Put<br /><br />PutxinEquation(1)in Equation \left(1\right), we get

(b+1b−1)2+(b+1b−1)+b=0{\left(\dfrac{b + 1}{b - 1}\right)}^{2} + \left(\dfrac{b + 1}{b - 1}\right) + b = 0

⇒⇒ (b+1)2+(b+1)(b−1)+b(b−1)2=0{\left(b + 1\right)}^{2} + \left(b + 1\right)\left(b - 1\right) + b{\left(b - 1\right)}^{2} = 0

⇒⇒ b2+1+2b+b2−1+b(b2−2b+1)=0{b}^{2} + 1 + 2b + {b}^{2} - 1 + b\left({b}^{2} - 2b + 1\right) = 0

⇒⇒ 2b2+2b+b3−2b2+b=02{b}^{2} + 2b + {b}^{3} - 2{b}^{2} + b = 0

⇒⇒ b3+3b=0{b}^{3} + 3b = 0
⇒⇒ b(b2+3)=0⇒b=0orb\left({b}^{2} + 3\right) = 0 ⇒ b = 0 or{b}^{2} = - 3<br/><br/>When<br /><br />Whenb = 0thencommonrootthen common rootx = - 1<br/>Butthegivencommonroot<br />But the given common rootx \ne - 1<br/><br/><br /><br />∴$b2=−3{b}^{2} = - 3

b=±3ib = \pm \sqrt{3}i

\left|b\right| = \sqrt{3}$
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