0
$\begingroup$

If $\alpha, \beta$ are roots of the equation $x^2 + px - q = 0$. $\gamma , \delta$ are roots of equation $x^2 + px -r$, then find the value of $(\alpha - \gamma )(\alpha - \delta)$.

Answer - $q-r$


My try -

$\alpha + \beta= -p$ and $\alpha \beta = -q$

similarly,

$\gamma + \delta = -p$ and $\gamma \delta = -r$

Then to we've to find:

$(\alpha - \gamma)(\alpha - \delta) = \alpha^2 - \alpha \delta - \alpha \gamma + \gamma \delta $ out of which only $\gamma \delta$ is known, then how to find the rest?

Also, when noticed carefully about the question, we find that question is $(\alpha - \gamma)(\alpha - \delta)$ which doesn't have $\beta$ in its product, which makes the question more confusing.

Thanks in Advance :)

$\endgroup$
  • $\begingroup$ Where is $\beta$ in your product? $\endgroup$ – Dr. Sonnhard Graubner Aug 19 '18 at 14:27
  • $\begingroup$ Isn't it $$(\alpha-\gamma)(\beta-\delta)$$? $\endgroup$ – Dr. Sonnhard Graubner Aug 19 '18 at 14:29
  • $\begingroup$ @Dr.SonnhardGraubner which product? Can you clarify a bit please? $\endgroup$ – Abhas Kumar Sinha Aug 19 '18 at 14:29
  • 1
    $\begingroup$ @Dr.SonnhardGraubner No, sir, if $\beta$ had been given in the equation, then I'd have solved the question, I again checked the question but didn't found any &\beta& $\endgroup$ – Abhas Kumar Sinha Aug 19 '18 at 14:30
  • 1
    $\begingroup$ You might as well leave the question, now. Pick whichever answer you like the most and accept that as your answer by clicking the tick, and then let the question stay. $\endgroup$ – Matt Aug 19 '18 at 15:14
2
$\begingroup$

We have $\alpha + \beta = \gamma + \delta \implies \beta = \gamma + \delta - \alpha$

Now, $(\alpha - \gamma)(\alpha - \delta) = \alpha^2 - \alpha \delta - \alpha \gamma + \gamma \delta = \alpha(\alpha - \delta - \gamma) + \gamma\delta = \alpha(-\beta) +\gamma\delta = \gamma\delta - \alpha\beta $

$= -r + q$

$\endgroup$
  • $\begingroup$ Seems like that there's misprint in textbook $\endgroup$ – Abhas Kumar Sinha Aug 19 '18 at 14:47
  • $\begingroup$ $(\alpha - \gamma)(\alpha - \delta) = 0$ iff there is a solution satifying both equations, which can only happen if q=r. Hence answer must be k(q-r) for some k. $\endgroup$ – dm63 Aug 19 '18 at 15:19
  • $\begingroup$ I don't understand what you're getting at here @dm63 $\endgroup$ – Matt Aug 19 '18 at 15:40
  • $\begingroup$ @Matt, I was looking at the possibility that $(\alpha-\gamma)(\alpha-\delta)=0$ This happens if and only if either $\alpha = \gamma$ or $\alpha= \delta$, which means that the equations have a common solution. Hence, we attempt to solve them simultaneously. Subtracting them gives q=r. Hence $(\alpha-\gamma)(\alpha-\delta)=0$ if and only if q=r, from which I concluded that $(\alpha-\gamma)(\alpha-\delta)=k(q-r)$ for some k. This is not needed , since the question has already been answered, but I though it might be interesting. I could not find a way of deducing k=-1 to complete the answer. $\endgroup$ – dm63 Aug 19 '18 at 18:45
1
$\begingroup$

We get by solving the quadratic $$x_{1,2}=-\frac{p}{2}\pm\sqrt{\frac{p^2}{4}+q}$$ and

$$x_{3,4}=-\frac{p}{2}\pm\sqrt{\frac{p^2}{4}+r}$$ so we get

$$(\alpha-\gamma)(\beta-\delta)=\left(\sqrt{\frac{p^2}{4}+q}-\sqrt{\frac{p^2}{4}+r}\right)\left(\sqrt{\frac{p^2}{4}+q}+\sqrt{\frac{p^2}{4}+r}\right)=\frac{p^2}{4}+q-\frac{p^2}{4}-r=q-r$$ as you stated.

$\endgroup$
  • $\begingroup$ Thanks for Clarification, the question is misprinted in the textbook $\endgroup$ – Abhas Kumar Sinha Aug 19 '18 at 14:44
1
$\begingroup$

By the given: $$(\alpha-\gamma)(\alpha-\delta)=\alpha^2+p\alpha-r=q-r.$$

$\endgroup$
  • $\begingroup$ Seems like misprint in my textbook answer, apologies $\endgroup$ – Abhas Kumar Sinha Aug 19 '18 at 14:48
  • $\begingroup$ @Abhas Kumar Sinha I did not see your answer. I also think so. $\endgroup$ – Michael Rozenberg Aug 19 '18 at 14:50
  • $\begingroup$ Should I edit the question or add details the the question might be incorrect? $\endgroup$ – Abhas Kumar Sinha Aug 19 '18 at 14:51
  • $\begingroup$ I think the idea of the solution would be the same idea. $\endgroup$ – Michael Rozenberg Aug 19 '18 at 14:53
  • $\begingroup$ which solution? $\endgroup$ – Abhas Kumar Sinha Aug 19 '18 at 14:54

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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