Question

find the value of x² - 1/x²​

Answer 1

$\star\:\:\:\bf\large\underline\blue{Question:-}$

• If $x=\frac{1}{2-\sqrt{3}}$, find the value of $x^{2}-\frac{1}{x^{2}}$.

$\star\:\:\:\bf\large\underline\blue{Solution:-}$

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

Therefore,

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

Now,

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

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

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

$= \frac{2 + \sqrt{3} }{4 - 3}$

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

$= 2 + \sqrt{3}$

So,

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

$= 4$

Also,

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

$= 2 \sqrt{3}$

So,

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

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

$= 4 \times 2 \sqrt{3}$

$= 8 \sqrt{3}$

$\star\:\:\:\bf\large\underline\blue{Answer:-}$

• ${x}^{2} - \frac{1}{ {x}^{2} } = 8 \sqrt{3}$