Question

if X = 8 + root 60 then root X + 1 / root X = ?

Answer 1

x=8+√60
=8+√4×15
=5+3+2√15
=(√5)²+2.√5.√3+(√3)²
=(√5+√3)²
∴, √x=√5+√3
1/√x=1/(√5+√3)
=(√5-√3)/(√5+√3)(√5-√3)
=(√5-√3)/{(√5)²-(√3)²}
=(√5-√3)/(5-3)
=(√5-√3)/2
√x+1/√x
=√5+√3+(√5-√3)/2
=(2√5+2√3+√5-√3)/2
=(3√5+√3)/2
=√3(√3.√5+1)/2
=√3(1+√15)/2

Answer 2

Answer:

$\sqrt{x}+\frac{1}{\sqrt{x}}=\frac{3\sqrt{5}+\sqrt{3}}{2}$

Step-by-step explanation:

Given : $x=8+\sqrt{60}$

To find : $\sqrt{x}+\frac{1}{\sqrt{x}}$

Solution :

First we find the square root of x,

$x=8+\sqrt{60}$

$x=8+\sqrt{4\times 15}$

$x=5+3+2\sqrt{15}$

$x=(\sqrt{5})^2+(\sqrt{3})^2+2\times \sqrt{5}\times \sqrt{3}$

$x=(\sqrt{5}+\sqrt{3})^2$

$\sqrt{x}=\sqrt{5}+\sqrt{3}$

Now, we find $\frac{1}{\sqrt{x}}$

$\frac{1}{\sqrt{x}}=\frac{1}{\sqrt{5}+\sqrt{3}}$

Rationalize,

$\frac{1}{\sqrt{x}}=\frac{1}{\sqrt{5}+\sqrt{3}}\times \frac{\sqrt{5}-\sqrt{3}}{\sqrt{5}-\sqrt{3}}$

$\frac{1}{\sqrt{x}}=\frac{\sqrt{5}-\sqrt{3}}{(\sqrt{5})^2-(\sqrt{3})^2}$

$\frac{1}{\sqrt{x}}=\frac{\sqrt{5}-\sqrt{3}}{5-3}$

$\frac{1}{\sqrt{x}}=\frac{\sqrt{5}-\sqrt{3}}{2}$

Substitute in the expression,

$\sqrt{x}+\frac{1}{\sqrt{x}}=\sqrt{5}+\sqrt{3}+\frac{\sqrt{5}-\sqrt{3}}{2}$

$\sqrt{x}+\frac{1}{\sqrt{x}}=\frac{2\sqrt{5}+2\sqrt{3}+\sqrt{5}-\sqrt{3}}{2}$

$\sqrt{x}+\frac{1}{\sqrt{x}}=\frac{3\sqrt{5}+\sqrt{3}}{2}$

Therefore, $\sqrt{x}+\frac{1}{\sqrt{x}}=\frac{3\sqrt{5}+\sqrt{3}}{2}$