Question

If $x=9+4\sqrt{5}$ then $\sqrt{x}-\frac{1}{\sqrt{x}}$is equal to?

Answer 1

$x=$$9+$$4\sqrt{5}$$=4+5+$$4\sqrt{5}$

$=$$(2{)}^2+{\left(\sqrt{5}\right)}^2+2(2\left)\right(\sqrt{5})$

$=$ $({2+\sqrt{5})}^2$

$=$ $\sqrt{x}-\frac{1}{\sqrt{x}}$

$=$ $\sqrt{{(2+\sqrt{5})}^2}-\frac{1}{\sqrt{{(2+\sqrt{5})}^2}}$

$=$ $2+\sqrt{5}-\frac{1}{2+\sqrt{5}}$

$=$ $2+\sqrt{5}-\frac{1}{2+\sqrt{5}}\times \frac{2-\sqrt{5}}{2-\sqrt{5}}$

$=$ $2+\sqrt{5}-\frac{2-\sqrt{5}}{2^2-{\sqrt{5}}^2}$

$=$ $2+\sqrt{5}-\frac{2-\sqrt{5}}{4-5}$

$=$ $2+\sqrt{5}-\frac{2-\sqrt{5}}{-1}$

$=$ $2+\sqrt{5}+\frac{2-\sqrt{5}}{1}$

$=$ $2+\sqrt{5}+2-\sqrt{5}$

$=$ $4$