Search any question & find its solution
Question:
Answered & Verified by Expert
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
Options:
Solution:
1453 Upvotes
Verified Answer
The correct answer is:
$\frac{2}{9}(2 p-q)(2 q-p)$
$\frac{2}{9}(2 p-q)(2 q-p)$
The equation $x^2-p x+r=0$ has roots $(\alpha, \beta)$ and the equation $x^2-q x+r=0$ has roots
$$
\begin{array}{ll}
\left(\frac{\alpha}{2}, 2 \beta\right) . & \\
\Rightarrow & r=\alpha \beta \text { and } \alpha+\beta=p \text { and } \frac{\alpha}{2}+2 \beta=q \\
\Rightarrow & \beta=\frac{2 q-p}{3} \text { and } \alpha=\frac{2(2 p-q)}{3} \\
\Rightarrow & \alpha \beta=r=\frac{2}{9}(2 q-p)(2 p-q)
\end{array}
$$
$$
\begin{array}{ll}
\left(\frac{\alpha}{2}, 2 \beta\right) . & \\
\Rightarrow & r=\alpha \beta \text { and } \alpha+\beta=p \text { and } \frac{\alpha}{2}+2 \beta=q \\
\Rightarrow & \beta=\frac{2 q-p}{3} \text { and } \alpha=\frac{2(2 p-q)}{3} \\
\Rightarrow & \alpha \beta=r=\frac{2}{9}(2 q-p)(2 p-q)
\end{array}
$$
Looking for more such questions to practice?
Download the MARKS App - The ultimate prep app for IIT JEE & NEET with chapter-wise PYQs, revision notes, formula sheets, custom tests & much more.