Question

$if x = 3 + 2 \sqrt{2} \: find \: the \: value \: of \: {x}^{2} \times \frac{1}{ {x}^{2} }$
plz answer
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Answer 1

Solution :-

I think there is a mistake in the question it should be :-

If x = 3 + 2√2 , find out the value of x² + 1/x² .

Now as the value of x = 3 + 2√2

Hence as 1/x = 1/(3 + 2√2)

Now by rationalising denominator :-

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

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

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

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

Now as a² + b² = (a + b)² - 2ab

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

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

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

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

$\implies x^2 + \dfrac{1}{x^2} = 34$