Two quadratic equations with equal ratios of roots 
If the ratio of roots of $ax^2+bx+c = 0\space$and $px^2+qx+r = 0\space$is same.
  How to find ratio of their discriminants?

I don't understand this problem,what exactly is meant by ratio of the roots being same?
Let, $\alpha, \beta$ and $\gamma,\delta$ are the roots of the two equations respectively,does this problem says that $\frac{\alpha}{\beta} = \frac{\gamma}{\delta}=k$, where $k \in \mathbb{Q}$?
Even so I am not really much ideas how to continue without messing with tedious algebraic manipulation,again,considering this problem is of quantitative aptitude category,it may not be the right approach.Any ideas?
 A: The ratio of the roots of the first quadratic polynomial are $(|b|-\sqrt{b^2-4ac})/(|b|+\sqrt{b^2-4ac})$. This is a one-to-one function of $ac/b^2$ hence the ratios coincide for the two polynomials if and only if $ac\cdot q^2=pr\cdot b^2$ $(*)$. 
The discriminants of the quadratic forms are $D=b^2-4ac$ and $\Delta=q^2-4pr$ hence $(*)$ is equivalent to the condition that $b^2\cdot\Delta=q^2\cdot D$.
A: you can use Vieta's formulas  :
$x_1+x_2=\frac{-b}{a}$  ,and $x_1x_2=\frac{c}{a}$  ,so we may write following:
$x_1^2+2x_1x_2+x_2^2=\frac{b^2}{a^2}$ , if we devide this equation by $x_1x_2$ we get
$\frac{x_1}{x_2}+\frac{x_2}{x_1}+2=\frac{b^2}{ac} \Rightarrow\frac{x_1}{x_2}+\frac{x_2}{x_1}-2=\frac{\Delta_1}{ac} \Rightarrow k+\frac{1}{k}-2=\frac{\Delta_1}{ac}$
Similarly we can show that $k+\frac{1}{k}-2=\frac{\Delta_2}{pr}$ ,so if you devide these last two equations you get:
$\frac{\Delta_1}{\Delta_2}=\frac{ac}{pr}$
A: let equations be $ax^2+bx+c=0$ and $px^2+qx+r=0$
let roots of $ax^2+bx+c=0$ be $m$ and $n$
let roots of $px^2+qx+r=0$ be $l$ and $o$
then ${m\over n}={l\over o}$
by componendo and dividendo law
${m+n\over m-n} ={l+o\over l-o}$
by solving them we get ${b^2-4ac\over q^2-4pr}={b^2\over q^2}$
