Question

Quardratic surds
If $x=\sqrt{\frac{\sqrt{5}+1}{\sqrt{5}+1}}$
Then prove that $x^2-x-1=0$

Answer 1

$x=\sqrt{\frac{\sqrt{5}+1}{\sqrt{5}+1}}x=\sqrt{\frac{(\sqrt{5}+1)(\sqrt{5}+1)}{\sqrt{5}-1}(\sqrt{5}-1)}=\sqrt{\frac{(\sqrt{5}+1{)}^2}{5-1}}=\frac{\sqrt{5}+1}{2}x=\frac{\sqrt{5}+1}{2}{x}^2=\frac{5+1+2\sqrt{5}}{4}=\frac{3+\sqrt{5}}{2}{x}^2-x-1=\frac{3}{2}+\frac{\text{/}\sqrt{5}}{\text{/}2}-\frac{\text{/}\sqrt{5}}{\text{/}2}-\frac{1}{2}-1=0$