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JEE Main 201322 AprilOnline
Exam
JEE Main
Year
2013
Date
22 April
Mode
Online
Subject
Mathematics
Chapter
Quadratic Equation

JEE Main 2013 · 22 April · Online · Mathematics · Quadratic Equation

If α and β are roots of the equation x^2+p x+(3 p)/(4)=0 , such...

If α\alpha and β\beta are roots of the equation x2+px+3p4=0x^2+p x+\frac{3 p}{4}=0, such that ∣α−β∣=10|\alpha-\beta|=\sqrt{10}, then pp belongs to the set :

Options

  1. A{2,−5}\{2,-5\}
  2. B{−3,2}\{-3,2\}
  3. C{−2,5}\{-2,5\}
  4. D{3,−5}\{3,-5\}

Correct answer

C. {−2,5}\{-2,5\}

Step-by-step solution

Given quadratic eqn. is x²+p x+3p4=\frac{3 p}{4}=0 So, α+β=−p,αβ=3p4\alpha+\beta=-p, \alpha \beta=\frac{3 p}{4} Now, given ∣α−β∣=10|\alpha-\beta|=\sqrt{10}⇒α−β=±10⇒(α−β)2=10⇒α2+β2−2αβ=10⇒(α+β)2−4αβ=10⇒p2−4×3p4=10⇒p2−3p−10=0⇒p=−2,5⇒p∈{−2,5}\begin{aligned}& \Rightarrow \alpha-\beta=\pm \sqrt{10}\\ & \Rightarrow(\alpha-\beta)^2=10 \Rightarrow \alpha^2+\beta^2-2 \alpha \beta=10 \\ & \Rightarrow(\alpha+\beta)^2-4 \alpha \beta=10 \\ & \Rightarrow p²-4 \times \frac{3 p}{4}=10 \Rightarrow p²-3 p-10=0 \\ & \Rightarrow p=-2,5 \Rightarrow p \in\{-2,5\} \end{aligned}$
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