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JEE Advanced · Mathematics · 2. Quadratic Equations

Let \(\alpha, \beta\) be the roots of the equation \(x^2-p x+r=0\) and \(\frac{\alpha}{2}, 2 \beta\) be the roots of the equation \(x^2-q x+r=0\). Then, the value of \(r\) is

  1. A \(\frac{2}{9}(p-q)(2 q-p)\)
  2. B \(\frac{2}{9}(q-p)(2 p-q)\)
  3. C \(\frac{2}{9}(q-2 p)(2 q-p)\)
  4. D \(\frac{2}{9}(2 p-q)(2 q-p)\)
Verified Solution

Answer & Solution

Correct Answer

(D) \(\frac{2}{9}(2 p-q)(2 q-p)\)

Step-by-step Solution

Detailed explanation

The equation \(x^2-p x+r=0\) has roots \((\alpha, \beta)\) and the equation \(x^2-q x+r=0\) has roots
\(\left(\frac{\alpha}{2}, 2 \beta\right) . \)
\( \Rightarrow r=\alpha \beta \text { and } \alpha+\beta=p \text { and }\)\(\frac{\alpha}{2}+2 \beta=q \)
\( \Rightarrow \beta=\frac{2 q-p}{3} \text { and } \alpha=\frac{2(2 p-q)}{3} \)
\( \Rightarrow \alpha \beta=r=\frac{2}{9}(2 q-p)(2 p-q)\)
From JEE Advanced
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