Question

If x= 2+ √3 then the value of √x + 1/√x is?

Answer 1

$\underline{\large{Answer\: \colon}}$

Given : x = 2 + √3

To Find :

$\sqrt{x} \: + \: \dfrac{1}{\sqrt{x}} \: = \: ?$

Solution :

x = 2 + √3

then $\dfrac{1}{x}$ is $\dfrac{1}{2\: + \: \sqrt{3}}$

Rationalise the value as given below ( follow the steps )

$\dfrac{1}{x}\: = \: \dfrac{1}{2\: +\: \sqrt{3}}\times \dfrac{2\: - \:\sqrt{3}}{2\: - \:\sqrt{3}}$

then ,

$\dfrac{1}{x} \: = \: \dfrac{2\: - \: \sqrt{3}}{2^{2} \: - \: {\sqrt{3}}^{2}}$

after rationalise we get the value of $\dfrac{1}{x}$ is (2 - √3)

$\dfrac{1}{x} \: = \: \dfrac{2\: - \: \sqrt{3}}{4\: - \: 3}$

$\dfrac{1}{x} \: = \: \dfrac{2\: - \: \sqrt{3}}{1}$

$\dfrac{1}{x} \: = \: (2\: - \: \sqrt{3})$

after that

$x \: + \: \dfrac{1}{x} \: = \: (2 \: + \sqrt{3}) \: + \: (2 \: - \: \sqrt{3})$

eliminate √3

$x \: + \: \dfrac{1}{x} \: = \: 2 \: + \: 2$

$x \: + \: \dfrac{1}{x} \: = \: 4$

To get the value of $\sqrt{x} \: + \: \dfrac{1}{\sqrt{x}}$ add " 2 " both side :

$x \: + \: \dfrac{1}{x} \: + \: 2= \: 4 \: + \: 2$

then

${(\sqrt{x}})^{2} + {\Big(\dfrac{1}{\sqrt{x}}\Big)^{2}} + 2 \times \sqrt{x} \times \dfrac{1}{\sqrt{x}} \: = \: 6$

${\Big(\sqrt{x}\: + \: \dfrac{1}{\sqrt{x}}\Big)}^{2} \: = \: 6$

$\sqrt{x} + \dfrac{1}{\sqrt{x}}\: = \: \sqrt{6}$

Hence answer is √6