Question

if x = √3+√2 then find x+1/x​

Answer 1

Answer :-

→ 2√3 .

Step-by-step explanation :-

We have ,

→ x = √3 + √2 .

Then,

$\sf \rightarrow \frac{1}{x} = \frac{1}{ \sqrt{3} + \sqrt{2} } . \\ \\ \sf = \frac{1}{ \sqrt{3} + \sqrt{2} } \times \frac{ \sqrt{3} - \sqrt{2} }{ \sqrt{3} - \sqrt{2} } . \\ \\ \sf= \frac{ \sqrt{3} - \sqrt{2} }{ {( \sqrt{3}) }^{2} - {( \sqrt{2}) }^{2} } . \\ \\ = \frac{ \sqrt{3} - \sqrt{2} }{3 - 2} . \\ \\ \it \therefore \frac{1}{x} = \sqrt{3} - \sqrt{2} .$

Now,

→ x + 1/x .

= √3 + √2 + √3 - √2 .

= 2√3 .

Hence , it is solved .

Answer 2

$\large\sf{given \: x = \sqrt{3} + \sqrt{2} }$

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

$\large\sf{rationalise \: the \: denominator}$

$\large\implies \sf{ \frac{1}{x} = \frac{1}{ \sqrt{3} + \sqrt{2} } \times \frac{ \sqrt{3} - \sqrt{2} }{ \sqrt{3} - \sqrt{2} } }$

$\large\implies \sf{ \frac{1}{x} = \frac{ \sqrt{3} - \sqrt{2} }{{( \sqrt{3})}^{2} - {( \sqrt{2})}^{2} } }$

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

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

now

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

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