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

$$\left|\left(r-3\frac{r+\sqrt{r^2-4}}{2}\right)\left(r-3\frac{r-\sqrt{r^2-4}}{2}\right)\right|=|9-2r^2|$$

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

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HINT:

$$(r+\sqrt{r^2-4})(r-\sqrt{r^2-4})=4$$

and

$$(r+\sqrt{r^2-4})+(r-\sqrt{r^2-4})=2r$$

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

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

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  • $\begingroup$ @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
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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.

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