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NDA 2016
Exam
NDA
Year
2016
Subject
Mathematics
Chapter
Quadratic Equation

NDA 2016 · Mathematics · Quadratic Equation

Let α and β(α 0 and c < 0 . Consider the following: 1. β < -α 2. β...

Let α\alpha and β(α<β)\beta(\alpha < \beta) be the roots of the equation x2+bx+c=0\mathrm{x}^2+\mathrm{bx}+\mathrm{c}=0, where b>0\mathrm{b} > 0 and c<0\mathrm{c} < 0. Consider the following: 1. β<−α2.β<∣α∣\beta < -\alpha 2. \beta < |\alpha| Which of the above is/are correct?

Options

  1. A1 only
  2. B2 only
  3. CBoth 1 and 2
  4. DNeither 1 nor 2

Correct answer

C. Both 1 and 2

Step-by-step solution

Given quadratic equation, x2+bx+c=0x^2+b x+c=0 and roots are αandβ\alpha and \beta where α<β\alpha < \beta. Hence roots of given quadratic equation are β=−b+b2−4c2α=−b−b2−4c2(∵α<β)⇒−α=b+b2−4c2 and ∣α∣=b+b2−4c2∴β<−α and β<∣α∣ both are correct. \begin{aligned} & \beta=\frac{-b+\sqrt{b^2-4 c}}{2} \\ & \alpha=\frac{-b-\sqrt{b^2-4 c}}{2} \quad (\because \alpha < \beta)\\ & \Rightarrow-\alpha=\frac{b+\sqrt{b^2-4 c}}{2} \text{ and }|\alpha|=\frac{b+\sqrt{b^2-4 c}}{2} \\ & \therefore \beta < -\alpha \text{ and } \beta < |\alpha| \text{ both are correct. } \end{aligned}
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