If the ratio of the roots of the equation $x^2+px+q=0$ are equal to the ratio of the roots of the equation $x^2+bx+c=0$ , then prove that $p^2c=b^2q$
Let $\alpha \& \beta$ be the roots of first equation then $\alpha + \beta = -p \& \alpha \beta = q$ Let $ \gamma \& \delta$ be the roots of the other equation then $\gamma + \delta = -b ; \gamma \delta =c$ As per the question $ \frac{\alpha}{\beta}=\frac{\gamma}{\delta}$ How to proceed further.
Best Answer
As you pointed out, $\alpha+\beta=-p$ and $\alpha\beta=q$.
Divide $(\alpha+\beta)^2$ by $\alpha\beta$. This is allowed, since if one of the roots of each equation is $0$, the result holds trivially. We get $$\frac{\alpha}{\beta}+2+\frac{\beta}{\alpha}=\frac{p^2}{q}.$$ Similarly, $$\frac{\gamma}{\delta}+2+\frac{\delta}{\gamma}=\frac{b^2}{c}.$$ Since $\dfrac{\alpha}{\beta}=\dfrac{\gamma}{\delta}$, it follows that $\dfrac{p^2}{q}=\dfrac{b^2}{c}$.