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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?

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It just means $\frac{\alpha}{\beta} = \frac{\gamma}{\delta}$ or $\frac{\alpha}{\beta}= \frac{\delta}{\gamma}$, no reason to assume that the ration lies in $\mathbb{Q}$ a priori (you don't even have that the coefficients lie in $\mathbb{Q}$. –  Arturo Magidin Oct 12 '11 at 6:57
    
@Arturo Magidin:Aha,thanks :) you are right,I missed it completely by assuming the coefficients lie in $\mathbb{Q}$.However,for the rest,would you advice using the quadratic formula and then manipulating things? –  Quixotic Oct 12 '11 at 7:00

2 Answers 2

up vote 4 down vote accepted

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 $b^2\cdot\Delta=q^2\cdot D$.

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Well,I know the concept of one-to-one function (in terms of set-relation and mapping) but however I couldn't see how that's applying here?! :/ –  Quixotic Oct 12 '11 at 7:12
    
You know that $u(x)=u(y)$ and you want to deduce that $x=y$. This is not true in general (consider $u(x)=x^2$ on the real line) but this holds for one-to-one functions. –  Did Oct 12 '11 at 7:16

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:

$\frac{\Delta_1}{\Delta_2}=\frac{ac}{pr}$

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Please explain the step : $\frac{x_1}{x_2}+\frac{x_2}{x_1}+2=\frac{b^2}{ac} \Rightarrow\frac{x_1}{x_2}+\frac{x_2}{x_1}-2=\frac{\Delta_1}{ac}$ –  Quixotic Oct 12 '11 at 9:33
    
Also,the answer is $\frac{\Delta_1}{\Delta_2}=\frac{b^2}{q^2}$ –  Quixotic Oct 12 '11 at 9:34
    
$\frac{x_1}{x_2}+\frac{x_2}{x_1}-2+4=\frac{b^2}{ac} \Rightarrow \frac{x_1}{x_2}+\frac{x_2}{x_1}-2=\frac{b^2}{ac}-4=\frac{b^2-4ac}{ac}=\frac{\Del‌​ta_1}{ac}$ –  pedja Oct 12 '11 at 9:40
    
@FoolForMath,yes it is.. –  pedja Oct 12 '11 at 9:43
    
Sorry,but I can't see how $\frac{\Delta_1}{\Delta_2}=\frac{ac}{pr}$ becomes $\frac{b^2}{q^2}$ ... –  Quixotic Oct 12 '11 at 11:07

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