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Quadratic Equation and Inequalities question

2016 · 9 Apr · Shift 1 · Q27
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Quadratic Equation and Inequalities question

2016 · 9 Apr · Shift 1 · Q27

JEE MainMathematicsQuadratic Equation and InequalitiesMCQ+4 / −1
If the equations x2 + bx−1 = 0 and x2 + x + b = 0 have a common root different from −1, then ∣b∣\left| b \right|∣b∣ is equal to :
  1. A
    2\sqrt 22​
  2. B
    2
  3. C
    3
  4. D
    3\sqrt 33​
View written solutionFree

Correct answer: D

  1. Let the common root be α\alphaα, with α≠−1\alpha \neq -1α=−1.

    Since α\alphaα is a common root of x2+bx−1=0andx2+x+b=0,x^2+bx-1=0 \quad \text{and} \quad x^2+x+b=0,x2+bx−1=0andx2+x+b=0, we have: α2+bα−1=0...(1)\alpha^2+b\alpha-1=0 \quad ...(1)α2+bα−1=0...(1) α2+α+b=0...(2)\alpha^2+\alpha+b=0 \quad ...(2)α2+α+b=0...(2)

  2. Subtract (2) from (1): (α2+bα−1)−(α2+α+b)=0\left(\alpha^2+b\alpha-1\right)-\left(\alpha^2+\alpha+b\right)=0(α2+bα−1)−(α2+α+b)=0 bα−α−b−1=0b\alpha-\alpha-b-1=0bα−α−b−1=0 α(b−1)−(b+1)=0\alpha(b-1)-(b+1)=0α(b−1)−(b+1)=0 α(b−1)=b+1\alpha(b-1)=b+1α(b−1)=b+1

  3. Since the common root is different from −1-1−1, check the special case b=1b=1b=1:

    • If b=1b=1b=1, then equations become x2+x−1=0andx2+x+1=0,x^2+x-1=0 \quad \text{and} \quad x^2+x+1=0,x2+x−1=0andx2+x+1=0, which clearly do not have a common root. So b≠1b\neq 1b=1, and we can write α=b+1b−1.\alpha=\frac{b+1}{b-1}.α=b−1b+1​.

  4. Substitute this into one of the equations, say (2): α2+α+b=0.\alpha^2+\alpha+b=0.α2+α+b=0. So, (b+1b−1)2+b+1b−1+b=0.\left(\frac{b+1}{b-1}\right)^2+\frac{b+1}{b-1}+b=0.(b−1b+1​)2+b−1b+1​+b=0.

    Multiply by (b−1)2(b-1)^2(b−1)2: (b+1)2+(b+1)(b−1)+b(b−1)2=0.(b+1)^2+(b+1)(b-1)+b(b-1)^2=0.(b+1)2+(b+1)(b−1)+b(b−1)2=0.

  5. Expand each term: (b+1)2=b2+2b+1, (b+1)^2=b^2+2b+1,(b+1)2=b2+2b+1, (b+1)(b−1)=b2−1, (b+1)(b-1)=b^2-1,(b+1)(b−1)=b2−1, b(b−1)2=b(b2−2b+1)=b3−2b2+b. b(b-1)^2=b(b^2-2b+1)=b^3-2b^2+b.b(b−1)2=b(b2−2b+1)=b3−2b2+b.

    Adding: b2+2b+1+b2−1+b3−2b2+b=0b^2+2b+1+b^2-1+b^3-2b^2+b=0b2+2b+1+b2−1+b3−2b2+b=0 b3+3b=0b^3+3b=0b3+3b=0 b(b2+3)=0.b(b^2+3)=0.b(b2+3)=0.

  6. For real bbb, this gives b=0.b=0.b=0. Then the equations are x2−1=0andx2+x=0.x^2-1=0 \quad \text{and} \quad x^2+x=0.x2−1=0andx2+x=0. Their roots are:

    • First: x=±1x=\pm 1x=±1
    • Second: x=0,−1x=0,-1x=0,−1

    These have no common root different from −1-1−1. So this route suggests something inconsistent for real bbb.

  7. Use a cleaner method:

    If α\alphaα is a common root, then from α2+bα−1=0\alpha^2+b\alpha-1=0α2+bα−1=0 and α2+α+b=0,\alpha^2+\alpha+b=0,α2+α+b=0, subtracting gives α(b−1)=b+1....(3)\alpha(b-1)=b+1. \quad ...(3)α(b−1)=b+1....(3)

    Rearranging (3): b(α−1)=α+1b(\alpha-1)=\alpha+1b(α−1)=α+1 b=α+1α−1,α≠1.b=\frac{\alpha+1}{\alpha-1}, \quad \alpha\neq 1.b=α−1α+1​,α=1.

  8. Substitute this value of bbb into equation (2): α2+α+α+1α−1=0.\alpha^2+\alpha+\frac{\alpha+1}{\alpha-1}=0.α2+α+α−1α+1​=0. Multiply by (α−1)(\alpha-1)(α−1): α2(α−1)+α(α−1)+(α+1)=0.\alpha^2(\alpha-1)+\alpha(\alpha-1)+(\alpha+1)=0.α2(α−1)+α(α−1)+(α+1)=0.

    Expand: α3−α2+α2−α+α+1=0\alpha^3-\alpha^2+\alpha^2-\alpha+\alpha+1=0α3−α2+α2−α+α+1=0 α3+1=0\alpha^3+1=0α3+1=0 (α+1)(α2−α+1)=0. (\alpha+1)(\alpha^2-\alpha+1)=0.(α+1)(α2−α+1)=0.

    Given α≠−1\alpha\neq -1α=−1, we must have α2−α+1=0.\alpha^2-\alpha+1=0.α2−α+1=0.

  9. Now compute bbb from b=α+1α−1.b=\frac{\alpha+1}{\alpha-1}.b=α−1α+1​. We need ∣b∣|b|∣b∣.

    From α2−α+1=0\alpha^2-\alpha+1=0α2−α+1=0, the roots are α=1±i32.\alpha=\frac{1\pm i\sqrt{3}}{2}.α=21±i3​​.

    Take α=1+i32\alpha=\frac{1+i\sqrt{3}}{2}α=21+i3​​. Then α+1=3+i32,α−1=−1+i32.\alpha+1=\frac{3+i\sqrt{3}}{2}, \qquad \alpha-1=\frac{-1+i\sqrt{3}}{2}.α+1=23+i3​​,α−1=2−1+i3​​.

    Hence ∣b∣=∣α+1α−1∣=∣α+1∣∣α−1∣.|b|=\left|\frac{\alpha+1}{\alpha-1}\right|=\frac{|\alpha+1|}{|\alpha-1|}.∣b∣=​α−1α+1​​=∣α−1∣∣α+1∣​.

    Now, ∣α+1∣=(32)2+(32)2=9+34=3,|\alpha+1|=\sqrt{\left(\frac{3}{2}\right)^2+\left(\frac{\sqrt{3}}{2}\right)^2}=\sqrt{\frac{9+3}{4}}=\sqrt{3},∣α+1∣=(23​)2+(23​​)2​=49+3​​=3​, ∣α−1∣=(−12)2+(32)2=1+34=1.|\alpha-1|=\sqrt{\left(-\frac{1}{2}\right)^2+\left(\frac{\sqrt{3}}{2}\right)^2}=\sqrt{\frac{1+3}{4}}=1.∣α−1∣=(−21​)2+(23​​)2​=41+3​​=1.

    Therefore, ∣b∣=3.|b|=\sqrt{3}.∣b∣=3​.

  10. So the correct option is 3.\boxed{\sqrt{3}}.3​​.

Thus, option D\boxed{D}D​ is correct.

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