Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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.

share|cite|improve this question

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 $\dfrac{p^2}{q}=\dfrac{b^2}{c}$.

share|cite|improve this answer
Divide $\alpha+\beta$ by $\alpha\beta$. We get $\frac{\alpha}{\beta}+\frac{\beta}{\alpha}=-\frac{p}{q}$... No we do not (and I am sure you will find the way so that we do). – Did Feb 20 '13 at 6:52
@Did: Thanks, wrong symmetric function. – André Nicolas Feb 20 '13 at 6:57
thanks a lot....all of you for support. – Sachin Sharmaa Feb 20 '13 at 8:47

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 $b=0\implies p^2c=0=b^2q$

share|cite|improve this answer
thanks a lot... – Sachin Sharmaa Feb 20 '13 at 16:53

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.