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

Solution 1:

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 $\boxed4$.

Solution 2:

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

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