Question

if x=2+√3 then find the value of
$x - \frac{1}{x}$
and also
$(x - \frac{1}{x} )^{2}$
​

Answer 1

Given

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

To Find

$\sf\to x-\dfrac{1}{x}$

$\sf\to\bigg( x-\dfrac{1}{x}\bigg)^2$

Now Take

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

$\sf\to \dfrac{1}{x} = \dfrac{1}{2+\sqrt{3} }$

Now using Rationalize the denominator

$\sf\to \dfrac{1}{x} = \dfrac{1}{2+\sqrt{3} }$

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

$\sf\to \dfrac{1}{x} = \dfrac{2-\sqrt{3} }{(2^2-(\sqrt{3})^2) }$

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

Now put the value

$\sf\to x-\dfrac{1}{x}$

We get

$\sf\to 2+\sqrt{3} -(2-\sqrt{3} )$

$\sf\to 2+\sqrt{3} -2+\sqrt{3}$

$\sf\to 2\sqrt{3}$

And

$\sf\to\bigg( x-\dfrac{1}{x}\bigg)^2$

We Have

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

$\sf\to\bigg( x-\dfrac{1}{x}\bigg)^2 = (2\sqrt{3} )^2= 4\times3=12$

Answer

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

$\sf\to\bigg( x-\dfrac{1}{x}\bigg)^2 =12$

Answer 2

Given :-

x = 2+√3

To Find :-

x - 1/x

$\sf\bigg(x - \dfrac{1}{x} \bigg)^{2}$

Solution :-

At first taking the reciprocal

$\bf \dfrac{1}{x} = \dfrac{1}{2 + \sqrt{3}}$

Now,

By rationalize the denominator

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

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

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

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

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

$\sf x = 2-\sqrt{3}$

Finding x - 1/x

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

$\sf = 2 - 2 + \sqrt{3} + \sqrt{3}$

$\sf 2\sqrt{3}$

$\tt \bigg\lgroup x - \dfrac{1}{x}\bigg\rgroup^{2} = (2\sqrt{3})^{2}$

$\tt \bigg\lgroup x - \dfrac{1}{x}\bigg\rgroup^{2} = 2^2 \times \sqrt{3}^{2}$

$\tt = 4 \times 3 =12$