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NDA 2017
Exam
NDA
Year
2017
Subject
Mathematics
Chapter
Quadratic Equation

NDA 2017 · Mathematics · Quadratic Equation

The sum of the roots of the equation x^2+b x+c=0 (where b and c are...

The sum of the roots of the equation x2+bx+c=0x^2+b x+c=0 (where b and c are non-zero) is equal to the sum of the reciprocals of their squares. Then 1c,b,cb\frac{1}{\mathrm{c}}, \mathrm{b}, \frac{\mathrm{c}}{\mathrm{b}} are in

Options

  1. AAP
  2. BGP
  3. CHP
  4. DNone of the above

Correct answer

C. HP

Step-by-step solution

Let α,β\alpha, \beta be the roots of x2+bx+c=0x^2+\mathrm{bx}+\mathrm{c}=0 Given, α+β=1α2+1β2\alpha+\beta=\frac{1}{\alpha^2}+\frac{1}{\beta^2}α+β=−b;α⋅β=c∴α+β=1α2+1β2⇒α+β=α2α2+β2β2=(α+β)2−2αβα2β2\begin{aligned} & \alpha+\beta=-\mathrm{b} ; \alpha \cdot \beta=\mathrm{c} \\ & \therefore \alpha+\beta=\frac{1}{\alpha^2}+\frac{1}{\beta^2} \\ & \Rightarrow \alpha+\beta=\frac{\alpha^2}{\alpha^2}+\frac{\beta^2}{\beta^2}=\frac{(\alpha+\beta)^2-2 \alpha \beta}{\alpha^2 \beta^2} \end{aligned}$\begin{aligned} & \Rightarrow-\mathrm{b}=\frac{\mathrm{b}^2-2 \mathrm{c}}{\mathrm{c}^2} \\ & \Rightarrow-\mathrm{bc}^2=\mathrm{b}^2-2 \mathrm{c} \Rightarrow 2 \mathrm{c}=\mathrm{b}^2+\mathrm{bc}^2 \\ & \Rightarrow \frac{2 \mathrm{c}}{\mathrm{ b}}=\mathrm{b}+\mathrm{c}^2 \Rightarrow \frac{2}{\mathrm{ b}}=\frac{\mathrm{b}}{\mathrm{c}}+\mathrm{c} \end{aligned} We know, if a,b,c\mathrm{a}, \mathrm{b}, \mathrm{c} are in A.P. 2b=a+c2 \mathrm{ b}=\mathrm{a}+\mathrm{c}. Similarly, we got 2b=bc+c\frac{2}{\mathrm{ b}}=\frac{\mathrm{b}}{\mathrm{c}}+\mathrm{c}, which means bc,1b,c\frac{\mathrm{b}}{\mathrm{c}}, \frac{1}{\mathrm{ b}}, \mathrm{c} are in A.P. So, cb,b,1c\frac{\mathrm{c}}{\mathrm{b}}, \mathrm{b}, \frac{1}{\mathrm{c}} are in H.P.
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