Question

If x=5+2√6,then show rhat √x + 1/√x=2√3

Answer 1

x=(√3)²+(√2)²+2.√3.√2
x=(√3+√2)²
√x=√3+√2

1/√x=1/√3+√2
1/√x=(1/√3+√2)*(√3-√2/√3-√2)
1/√x=√3-√2/(√3)²-(√2)²
1/√x=√3-√2/3-2
1/√x=√3-√2/1

√x+1/√x=√3+√2+√3-√2
=2√3

Answer 2

Answer:

Given,

$x=5+2\sqrt{6}$

$=5+2\sqrt{3\times 2}$

$=5+2\times \sqrt{3}\times \sqrt{2}$

$=3+2+2\times \sqrt{3}\times \sqrt{2}$

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

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

Thus,

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

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

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

$=\frac{\sqrt{3}-\sqrt{2}}{3-2}$

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

$=\sqrt{3}-\sqrt{2}$

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

Hence, proved....