Question

if x = 5+2√6 then find √x+1/√x​

Answer 1

Answer:

$2\sqrt{3}$

Step-by-step explanation:

Given :

To find the value of , $\sqrt{x} + \frac{1}{\sqrt{x} }$ if $x = 5 + 2\sqrt{6}$

Solution :

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

⇒$x = 3 + 2 + 2\sqrt{6}$

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

⇒$x = (\sqrt{3} + \sqrt{2})^2$

Hence,

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

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

By taking conjugate, We get,

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

$= \frac{\sqrt{3}-\sqrt{2}}{(\sqrt{3} + \sqrt{2})(\sqrt{3} - \sqrt{2})}$

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

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

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

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

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

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