Question

if x=2-√3,then find value of x^2 - 1÷x^2

Answer 1

Given that,

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

Now,

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

To find the value of,

$\sf{x {}^{2} - \frac{1}{x {}^{2} } }$

It is of the form,

a² - b²=(a+b)(a-b)

Here,

$\sf{a = x \: \: and \: \: b = \frac{1}{x} }$

Putting the values,

x²-1/x²

=(x+1/x)(x-1/x)

=[(2- √3)+(2+√3)][(2- √3)-(2+√3)]

=(4)(-2√3)

= -8√3

The required value is -8√3