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Question: Express the roots of the equation $q^2x^2-(p^2-2q)x+1=0$ in terms of those of $x^2+px+q=0$

My attempt:

Roots of the second equation are $\frac{-p±\sqrt{p^2-4q}}{2}$

Roots of the first equation are $\frac{(p^2-2q)±\sqrt{(p^2-2q)^2-4q^2}}{2q^2}$ which on simplification becomes $\frac{(p^2-2q)±p\sqrt{(p^2-4q)}}{2q^2}$

The discriminants of both the equations are the same.

My problem: I am unable to proceed further. I was trying to take the ratio of the roots and find a relation between them but this is not working out. I am also unsure whether the logic that i have used in my attempt is of any use.

MrAP
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    Let $x_1,x_2$ be the roots of the second equation, then $p=-(x_1+x_2)$ and $q=x_1x_2$ by Vieta's relations. Replace that in your last formula and simplify. – dxiv Jun 17 '17 at 02:00

1 Answers1

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Let the root of the second equation to $\alpha , \beta$

Then $$\alpha + \beta = -p$$

$$\alpha \beta = q$$

$$\frac{p^2-2q}{q^2}=\frac{(\alpha+\beta)^2-2\alpha\beta}{(\alpha\beta)^2}=\frac{\alpha^2+\beta^2}{(\alpha\beta)^2}=\frac{1}{\beta^2}+\frac{1}{\alpha^2}$$

$$\frac{1}{q^2}=\frac{1}{\alpha^2}\frac{1}{\beta^2}$$

Can you see the root for the first equation now?

Siong Thye Goh
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