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NDA 2019
Exam
NDA
Year
2019
Subject
Mathematics
Chapter
Quadratic Equation

NDA 2019 · Mathematics · Quadratic Equation

If the roots of the equation x^2+p x+q=0 are 19^ ^ and 26^ ^ , then...

If the roots of the equation x2+px+q=0x^2+p x+q=0 are tan⁡19∘\tan 19^{{}^{\circ}} and tan⁡26∘\tan 26^{{}^{\circ}}, then which one of the following is correct ?

Options

  1. Aq−p=1\mathrm{q}-\mathrm{p}=1
  2. Bp−q=1\mathrm{p}-\mathrm{q}=1
  3. Cp+q=2p+q=2
  4. Dp+q=3p+q=3

Correct answer

A. q−p=1\mathrm{q}-\mathrm{p}=1

Step-by-step solution

Given, tan⁡19∘\tan 19^{{}^{\circ}} and tan⁡26∘\tan 26^{{}^{\circ}} are roots of x2+px+q=0∴tan⁡19∘+tan⁡26∘=−p1=−p\mathrm{x}^2+\mathrm{px}+\mathrm{q}=0\therefore \tan 19^{{}^{\circ}}+\tan 26^{{}^{\circ}}=\frac{-p}{1}=-p(tan⁡19∘)(tan⁡26∘)=q1=qtan⁡(19∘+26∘)=tan⁡19∘+tan⁡26∘1−tan⁡19∘tan⁡26∘tan⁡45∘=−p1−q⇒1=−p1−q⇒1−q=−p⇒q−p=1\begin{aligned} & \left(\tan 19^{{}^{\circ}}\right)\left(\tan 26^{{}^{\circ}}\right)=\frac{q}{1}=q \\ & \tan \left(19^{{}^{\circ}}+26^{{}^{\circ}}\right)=\frac{\tan 19^{{}^{\circ}}+\tan 26^{{}^{\circ}}}{1-\tan 19^{{}^{\circ}} \tan 26^{{}^{\circ}}} \\ & \tan 45^{{}^{\circ}}=\frac{-p}{1-q} \\ & \Rightarrow 1=\frac{-p}{1-q} \\ & \Rightarrow 1-\mathrm{q}=-\mathrm{p} \\ & \Rightarrow \mathrm{q}-\mathrm{p}=1\end{aligned}$$
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