Question

If x=7+4√3then find x power 2 - 1/x power 2

Answer 1

Given:-

$x = 7 + 4 \sqrt{3}$

To find :-

${x}^{2} - \dfrac{1}{ {x}^{2} }$

Solution:-

If x = 7 + 4√3

then

$\dfrac{1}{x} = \dfrac{1}{7 + 4 \sqrt{3} }$

$\dfrac{1}{x} = \dfrac{1}{7 + 4 \sqrt{3} } \times \dfrac{7 - 4 \sqrt{3} }{7 - 4 \sqrt{3} }$

$\dfrac{1}{x} = \dfrac{ 7 - 4 \sqrt{3} }{ {(7)}^{2} - {(4 \sqrt{3} )}^{2} }$

$\dfrac{1}{x} = \dfrac{7 - 4 \sqrt{3} }{49 - 48}$

$\dfrac{1}{x} = \dfrac{7 - 4 \sqrt{3} }{1}$

Now, we have to find,

${x}^{2} - \dfrac{1}{ {x}^{2} }$

It can also be written as :-

$(x + \dfrac{1}{x} ) (x - \dfrac{1}{x})$

Now, put the value of x and 1/x,

$= (7 + 4 \sqrt{3} + 7 - 4 \sqrt{3} )(7 + 4 \sqrt{3} - 7 + 4 \sqrt{3} )$

$= (14)(2 \times 4 \sqrt{3} )$

$= 14 \times 8 \sqrt{3}$

$= 112 \sqrt{3}$

hence, the required answer is 112√3.