Question

If x = 2 + V3, find the value of x2​

Answer 1

Step-by-step explanation:

x = 2+ √3

therefore

x square = (2+√3)(2+√3)

= 4 + 3 + 4√3

= 7 + 4√3

Answer 2

Correct question -

$If \: \: x = 2 + \sqrt{3} \:\: \: then \: \: find \: \: the \: \: values \: of \: {x}^{2} + \frac{1}{ {x}^{2} }$

Solution -

$x = 2 + \sqrt{3}$

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

Rationalising the denominator we will get -

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

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

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

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

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

Now,

$= {x}^{2} + \frac{1}{ {x}^{2} }$

Putting the values -

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

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

$= 4 + 3 + 4 \sqrt{3} + 4 + 3 - 4 \sqrt{3}$

$= 4 + 3 + 4 + 3$

$= 14$

$\therefore \: \: the \: \: value \: of \: {x}^{2} + \frac{1}{ {x}^{2} } \: is \: 14$