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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Answer:

Correct Answer: 32. (a)

Solution:

  1. 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.



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