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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.

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

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  • $\begingroup$ 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). $\endgroup$ – Did Feb 20 '13 at 6:52
  • $\begingroup$ @Did: Thanks, wrong symmetric function. $\endgroup$ – André Nicolas Feb 20 '13 at 6:57
  • $\begingroup$ thanks a lot....all of you for support. $\endgroup$ – Sachin Sharmaa Feb 20 '13 at 8:47
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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 $b=0\implies p^2c=0=b^2q$

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