Theory of Equations 1 Question 31
32. If $\alpha$ and $\beta$ are the roots of $x^{2}+p x+q=0$ and $\alpha^{4}, \beta^{4}$ are the roots of $x^{2}-r x+s=0$, then the equation $x^{2}-4 q x+2 q^{2}-r=0$ has always
(1989, 2M)
(a) two real roots
(b) two positive roots
(c) two negative roots
(d) one positive and one negative root
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Solution:
- Since, $\alpha, \beta$ are the roots of $x^{2}+p x+q=0$ and $\alpha^{4}, \beta^{4}$ are the roots of $x^{2}-r x+s=0$.
$\Rightarrow \alpha+\beta=-p ; \alpha \beta=q ; \alpha^{4}+\beta^{4}=r$ and $\alpha^{4} \beta^{4}=s$
Let roots of $x^{2}-4 q x+\left(2 q^{2}-r\right)=0$ be $\alpha^{\prime}$ and $\beta^{\prime}$
Now, $\quad \alpha^{\prime} \beta^{\prime}=\left(2 q^{2}-r\right)=2(\alpha \beta)^{2}-\left(\alpha^{4}+\beta^{4}\right)$
$$ =-\left(\alpha^{4}+\beta^{4}-2 \alpha^{2} \beta^{2}\right)=-\left(\alpha^{2}-\beta^{2}\right)^{2}<0 $$
$\Rightarrow$ Roots are real and of opposite sign.
