I am looking for some explanation and steps to show how the following expression was simplified to get the correct result.


I have been trying to figure out how they simplified this but have been stuck on the process. Any solution would be greatly appreciated






SPOLIER ALERT: Scroll over the highlighted area to reveal the solution.

Note that we have $$\begin{align}\left(r-3\frac{r+\sqrt{r^2-4}}{2}\right)\left(r-3\frac{r-\sqrt{r^2-4}}{2}\right)&=r^2\\\\&-\frac32 r\left((r+\sqrt{r^2-4})+(r-\sqrt{r^2-4})\right)\\\\&+\frac94(r+\sqrt{r^2-4})(r-\sqrt{r^2-4})\\\\&=r^2-\frac32r(2r)+\frac94(4)\\\\&=9-2 r^2\end{align}$$

  • $\begingroup$ @jh123 How does it not help? $\endgroup$ – Mark Viola Apr 12 '17 at 3:32
  • $\begingroup$ Would the foil method work in this situation? $\endgroup$ – jh123 Apr 12 '17 at 3:35
  • $\begingroup$ @jh123 Yes, it does. I've edited to add the solution in the highlighted area. $\endgroup$ – Mark Viola Apr 12 '17 at 3:38
  • $\begingroup$ @Dr.MV thanks for your submition! but in the last two steps shouldn't the answer be $9-2r^2?$ since $r^2-\frac{3}{2}r(2r)+\frac{4}{9}(4)=r^2-3r^2+9?$ $\endgroup$ – fr14 Apr 12 '17 at 9:32
  • $\begingroup$ Indeed. I've edited the typo. $\endgroup$ – Mark Viola Apr 12 '17 at 12:02

Another way to look at it is:

Take $3$s out of brackets: $9\left(\frac{r}{3}-\frac{r+\sqrt{r^2-4}}{2}\right)\left(\frac{r}{3}-\frac{r-\sqrt{r^2-4}}{2}\right)$

Now consider the quadratic factorization: $x^2-rx+1=\left(x-\frac{r+\sqrt{r^2-4}}{2}\right)\left(x-\frac{r-\sqrt{r^2-4}}{2}\right)$.

Multiplying both sides by $9$ and plugging $x=\frac{r}{3}$ will do.


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.