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NDA 2017
Exam
NDA
Year
2017
Subject
Mathematics
Chapter
Quadratic Equation

NDA 2017 · Mathematics · Quadratic Equation

If the roots of the equation x^2+p x+q=0 are in the same ratio as...

If the roots of the equation x2+px+q=0x^2+p x+q=0 are in the same ratio as those of the equation x2+lx+m=0x^2+l x+m=0, then which one of the following is correct?

Options

  1. Ap2m=l2q\mathrm{p}^2 \mathrm{ m}=l^2 \mathrm{q}
  2. Bm2p=l2p\mathrm{m}^2 \mathrm{p}=l^2 \mathrm{p}
  3. Cm2\mathrm{m}^2\mathrm{p}=\mathrm{q}2l^2 l
  4. Dm2p2=l2q\mathrm{m}^2 \mathrm{p}^2=l^2 \mathrm{q}

Correct answer

A. p2m=l2q\mathrm{p}^2 \mathrm{ m}=l^2 \mathrm{q}

Step-by-step solution

Let a,ba, b be the roots of x2+px+q=0x^2+p x+q=0 So, a+b=−p,ab=q\mathrm{a}+\mathrm{b}=-\mathrm{p}, \mathrm{ab}=\mathrm{q}....(i) Let c,d\mathrm{c}, \mathrm{d} be the roots of x2+ℓx+m=0\mathrm{x}^2+\ell \mathrm{x}+\mathrm{m}=0 So, c+d=−ℓ,cd=m\mathrm{c}+\mathrm{d}=-\ell, \mathrm{cd}=\mathrm{m}....(ii) Given that roots of both the equations are in the Same ratio. So, ab=cd\frac{\mathrm{a}}{\mathrm{b}}=\frac{\mathrm{c}}{\mathrm{d}}....(iii) ⇒ba=dc\Rightarrow \frac{\mathrm{b}}{\mathrm{a}}=\frac{\mathrm{d}}{\mathrm{c}}...(iv)  (iii) + (iv) ⇒ab+ba=cd+dc⇒a2+b2ab=c2+d2cd⇒a2+b2ab+2=c2+d2cd+2⇒a2+b2+2abab=c2+d2+2cdcd⇒(a+b)2ab=(c+d)2cd⇒(−P)2q=(−ℓ)2m (from (i) and (ii)) ⇒P2m=ℓ2q.\begin{aligned} & \text{ (iii) }+ \text{ (iv) } \Rightarrow \frac{\mathrm{a}}{\mathrm{b}}+\frac{\mathrm{b}}{\mathrm{a}}=\frac{\mathrm{c}}{\mathrm{d}}+\frac{\mathrm{d}}{\mathrm{c}} \\ & \Rightarrow \frac{\mathrm{a}^2+\mathrm{b}^2}{\mathrm{ab}}=\frac{\mathrm{c}^2+\mathrm{d}^2}{\mathrm{ cd}} \Rightarrow \frac{\mathrm{a}^2+\mathrm{b}^2}{\mathrm{ab}}+2=\frac{\mathrm{c}^2+\mathrm{d}^2}{\mathrm{ cd}}+2 \\ & \Rightarrow \frac{\mathrm{a}^2+\mathrm{b}^2+2 \mathrm{ab}}{\mathrm{ab}}=\frac{\mathrm{c}^2+\mathrm{d}^2+2 \mathrm{ cd}}{\mathrm{ cd}} \\ & \Rightarrow \frac{(\mathrm{a}+\mathrm{b})^2}{\mathrm{ab}}=\frac{(\mathrm{c}+\mathrm{d})^2}{\mathrm{ cd}} \\ & \Rightarrow \frac{(-\mathrm{P})^2}{\mathrm{q}}=\frac{(-\ell)^2}{\mathrm{ m}} \quad \text{ (from (i) and (ii)) } \\ & \Rightarrow \mathrm{P}^2 \mathrm{ m}=\ell^2 \mathrm{q} .\end{aligned}
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