Question

Find the square root of the following in the form of a binomial surd.
$n+\sqrt{n^2-1}$.

Answer 1

Let $x=\sqrt{n+\sqrt{n^2-1}}=\sqrt{x}+\sqrt{y}$

$n+\sqrt{n^2-1}=x+y+2\sqrt{xy}$

Hence, $x+y=n,xy=\frac{n^2-1}{4}$

$4x^2-4nx+n^2-1=0$

$x=\frac{(n+1)}{2}$ or $x=\frac{(n-1)}{2}$

Similarily, $y=\frac{(n-1)}{2}$ or $y=\frac{(n+1)}{2}$