Quadratic Equation relation between roots 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. 
 A: 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 $\dfrac{p^2}{q}=\dfrac{b^2}{c}$.
A: Let  $ \frac{\beta}{\alpha}=\frac{\delta}{\gamma}=t$ 
So, $$\alpha+\beta=-p\implies \alpha+\alpha t=-p\implies \alpha(1+t)=-p;
\alpha\beta=q\implies \alpha\cdot \alpha t=q$$ 
Similarly, and  $$\gamma(1+t)=-b,\gamma\cdot \gamma t=c$$
On division, $$\frac{ \alpha(1+t)}{\gamma(1+t)}=\frac pb \text{ and }\frac{\alpha\cdot \alpha t}{\gamma\cdot \gamma t}=\frac qc$$
or $$\frac  \alpha\gamma=\frac pb \text{ and }\frac{\alpha^2}{\gamma^2}=\frac qc \text{ if }t(t+1)\ne0$$
Then, equating values of $\left(\frac \alpha \gamma\right)^2, \frac qc=\left(\frac pb\right)^2\implies p^2c=b^2q$
If $t=0,q=0$ and $c=0\implies p^2c=0=b^2q$
If $t+1=0,t=-1,p=0$ and $b=0\implies p^2c=0=b^2q$
