# If $x=\sqrt{{\sqrt{5}+1\over \sqrt{5}-1}}$ What is the value of $5x^2-5x-1?$

If $$x=\sqrt{{\sqrt{5}+1\over \sqrt{5}-1}}$$ What is the value of $$5x^2-5x-1?$$

Efforts:

After rationalization, I got $$x={\sqrt{5}+1\over 2}$$ and $$x^2={\sqrt{5}+1\over \sqrt{5}-1}$$. Going by this method is very tedious and boring.

Is there a more clever and imaginative way to approach this problem?

Thanks.

• I guess once you recognise that $\frac{\sqrt5+1}{\sqrt5-1}$ is the square of $\frac{\sqrt5+1}2$, you can use the fact that $\frac{\sqrt5+1}2$ is a root of $x^2-x-1$. I don't know if this qualifies a clever. Some people have some love for qualifying knowledge by heart of facts about the golden ratio as clever, which may also be used to recognize that the fraction was a square. – Gae. S. Nov 16 '19 at 0:01

Rationalizing the denominator of $$x$$, we get $$x=\sqrt{\frac{\sqrt5+1}{\sqrt5-1}}=\sqrt{\frac{\sqrt5+1}{\sqrt5-1}\cdot\frac{\sqrt5+1}{\sqrt5+1}}=\sqrt{\frac{(\sqrt5+1)^2}{4}}=\frac{1+\sqrt5}2$$

For the expression to be evaluated, we have \begin{align} 5x^2-5x-1&=5(x^2-x-\frac15)\\ &=5(x^2-x-\frac15-\frac45+\frac45)\\ &=5(x^2-x-1)+4 \end{align}

Since $$x=\dfrac{1+\sqrt5}2$$ is a root of $$x^2-x-1=0$$, we have the expression to be equal to $$\boxed4$$.

You already have $$x=\frac{\sqrt5+1}{2}$$. Then $$x^2=\frac{\sqrt5+3}{2}$$ and so $$x^2=x+1$$.

Therefore $$5x^2-5x-1=5x+5-5x-1=4.$$

Hopeful I can reveal everybody's dirty little secret and concurrently avoid the fate of Hippasus of Metapontum.

The golden ratio ($$\phi = \frac{1+\sqrt{5}}{2}$$) is a well know proportion in mathematics. It's known to be the ratio of the length of a diagonal of a regular pentagon to one of it's edges. If you cut a string into two shorter lengths, such that the ratio of the length of the longer piece to that of the shorter piece is equal to the ratio of the original length of the string to that of the longer piece, then that ratio is $$\phi$$ (phi).

These aforementioned problems both generate the equation $$\phi^2-\phi-1 = 0$$, which we can then manipulate and express as $$\phi^2=\phi+1$$, or even as $$\frac{1}{\phi} = \phi - 1$$. Using these identities can make reducing expressions that include $$\sqrt{5}$$ and integers much easier. For example, in your problem, after finding that $$x = \phi$$, we could then apply the identity $$\phi-1 = \frac{1}{\phi}$$.

\begin{align} 5x^2 - 5x - 1 &= 5x(x-1) - 1\\ &= 5x\bigg(\frac{1}{x}\bigg) - 1\\ &= 5-1 = 4\\ \end{align}

Interestingly, if you think about it, the identity $$\phi^2 = \phi+1$$ may suggest a strategy for solving the Fibonacci sequence. e.g. $$\phi^n = \phi^{n-2}\phi^2=\phi^{n-2}(\phi+1) = \phi^{n-1} +\phi^{n-2}$$ but I digress.

You could also have used these identities in part 1 of the problem, namely ... \begin{align} x &= \sqrt\frac{\sqrt{5}+1}{\sqrt{5}-1}\\ &= \sqrt\frac{\sqrt{5}+1}{\sqrt{5}+1-2}\\ &= \sqrt\frac{\frac{\sqrt{5}+1}{2}}{\frac{\sqrt{5}+1-2}{2}}\\ &= \sqrt\frac{\phi}{\phi - 1}=\sqrt\frac{\phi}{\frac{1}{\phi}} = \sqrt{\phi^2} = \phi\\ \end{align}

With more experience, you'll gain better judgement in when you it's best to convert expressions with $$\sqrt{5}$$s to expressions with $$\phi$$s, and when doing so just creates more work for yourself.