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If $a= (5-\sqrt5)/(5+\sqrt5)$, evaluate, showing working:

  1. $a+\frac{1}{a}$
  2. $a^2+\frac{1}{a^2}$

I attempted these questions but got them wrong, the answer for part (1) is $3$, I got $2/5$.

My working:

\begin{align} a+1/a &= \frac{5-\sqrt5}{5+\sqrt5} + \frac{5+\sqrt5}{5-\sqrt5}\\ &=\frac{(5-\sqrt5)^2 + (5+\sqrt5)^2}{(5+\sqrt5)(5+\sqrt5)} \quad &\text{Common denominator}\\ &=\frac{(25-10\sqrt5 -5)+(25+10\sqrt5+5)}{25-5} \quad &\text{Expanding the brackets}\\ &=\frac{50}{20}\\ &=\frac{5}{2}\\ \end{align}

I've triple checked my working out, and can't find any algebraic mistakes. How is the answer $3$?

For question $2$, how is it done? I'm still a bit confused. Can the $a^2 +1/a^2$ be changed to something to do with part $1$?

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  • $\begingroup$ What about a^2+1/a^2? $\endgroup$ – Zach Apr 26 '18 at 0:32
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    $\begingroup$ $(-\surd 5)^2=+5$ so $(5-\surd 5)^2 = (25-10\surd 5+5)$ $\endgroup$ – Graham Kemp Apr 26 '18 at 0:40
  • $\begingroup$ I edited your question. Check out the following links to learn the basics of how to format, and you can always check my edit to see how I did it. math.meta.stackexchange.com/questions/5020/… $\endgroup$ – John Lou Apr 26 '18 at 0:41
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    $\begingroup$ @Zach For your second one, notice that $(a+ 1/a)^2 = a^2 + 1/a^2 + 2$ $\endgroup$ – John Lou Apr 26 '18 at 0:43
  • $\begingroup$ Thanks John, I'll look into it. $\endgroup$ – Zach Apr 26 '18 at 0:46
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You've made a mistake while expanding $(5-\sqrt5)^2$. It's $25-10\sqrt{5}+5.$ For the second part, use the fact that$a^{2}+\frac{1}{a^2}=(a+\frac{1}{a})^2-2$.

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  • $\begingroup$ Thanks, cleared it up for me $\endgroup$ – Zach Apr 26 '18 at 0:31
  • $\begingroup$ However, if the question was a^2 +a/a^2 would you write it all out or is there a better method? $\endgroup$ – Zach Apr 26 '18 at 0:33
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After getting a common denominator, you have $25 -20\sqrt{5}-5$ as the start of the numerator. The last term I wrote there should be $+5$, not $-5$.

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