Theory of Equations 1 Question 22
23. 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
$(2007,3 M)$
(a) $\frac{2}{9}(p-q)(2 q-p)$
(b) $\frac{2}{9}(q-p)(2 p-q)$
(c) $\frac{2}{9}(q-2 p)(2 q-p)$
(d) $\frac{2}{9}(2 p-q)(2 q-p)$
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Solution:
- The equation $x^{2}-p x+r=0$ has roots $\alpha, \beta$ and the equation $x^{2}-q x+r=0$ has roots $\frac{\alpha}{2}, 2 \beta$.
$$ \Rightarrow \quad r=\alpha \beta \text { and } \alpha+\beta=p $$
and $\frac{\alpha}{2}+2 \beta=q \Rightarrow \beta=\frac{2 q-p}{3}$ and $\alpha=\frac{2(2 p-q)}{3}$
$$ \Rightarrow \quad \alpha \beta=r=\frac{2}{9}(2 q-p)(2 p-q) $$
