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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Answer:

Correct Answer: 23. (d)

Solution:

  1. 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) $



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