Question

solve it pls........​

Answer 1

Answer:

hola mate

see the attachment.

hope it helps

Answer 2

Solution :-

As given that

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

Questions asked

Find out the value of

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

First of all we should rationalise the denominator of 1/x

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

By rationalising denominator

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

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

$\dfrac{1}{x} = \dfrac{3 + 2\sqrt{2}}{9 - 4(2)}$

$\dfrac{1}{x} = \dfrac{3 + 2\sqrt{2}}{9 - 8}$

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

Now as we know that

a² + b² = (a + b)² - 2ab

$x^2 + \dfrac{1}{x^2} = \left(x + \dfrac{1}{x}\right)^2 - 2 \times x \times \dfrac{1}{x}$

$x^2 + \dfrac{1}{x^2} = \left(3 - 2\sqrt{2} + 3 + 2\sqrt{2}\right)^2 - 2$

$x^2 + \dfrac{1}{x^2} = \left(6 \right)^2 - 2$

$x^2 + \dfrac{1}{x^2} = 36 - 2$

So

$\huge{\boxed{\sf{x^2 + \dfrac{1}{x^2} = 34 }}}$