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NDA 2015
Exam
NDA
Year
2015
Subject
Mathematics
Chapter
Quadratic Equation

NDA 2015 · Mathematics · Quadratic Equation

If the roots of the equation x^2-nx+m=0 differ by 1 , then

If the roots of the equation x2−nx+m=0\mathrm{x}^2-\mathrm{nx}+\mathrm{m}=0 differ by 1 , then

Options

  1. An2−4m−1=0\mathrm{n}^2-4 \mathrm{ m}-1=0
  2. Bn2+4m−1=0\mathrm{n}^2+4 \mathrm{ m}-1=0
  3. Cm2+4n+1=0\mathrm{m}^2+4 \mathrm{n}+1=0
  4. Dm2−4n−1=0\mathrm{m}^2-4 \mathrm{n}-1=0

Correct answer

A. n2−4m−1=0\mathrm{n}^2-4 \mathrm{ m}-1=0

Step-by-step solution

Let the root be α\alpha and β\beta∵x2−nx+m=0⇒α+β=n;α⋅β=m⇒α−β=1⇒(α+β)2=(α−β)2+4αβ⇒n2=1+4m⇒n2−4m−1=0\begin{array}{ll} \because & \mathrm{x}^2-\mathrm{nx}+\mathrm{m}=0 \\ \Rightarrow & \alpha+\beta=\mathrm{n} ; \alpha \cdot \beta=\mathrm{m} \\ \Rightarrow & \alpha-\beta=1 \\ \Rightarrow & (\alpha+\beta)^2=(\alpha-\beta)^2+4 \alpha \beta \\ \Rightarrow & \mathrm{n}^2=1+4 \mathrm{ m} \\ \Rightarrow & \mathrm{n}^2-4 \mathrm{ m}-1=0 \end{array}$$
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