Question

,If x=3-2√2 find x²+(1/x²)

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Answer 1

$\\ \large\underline{ \underline{ \sf{ \red{given:} }}}$

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

$\\ \\ \large\underline{ \underline{ \sf{ \red{to \: find:} }}}$

$\sf \: {x}^{2} + \frac{1}{ {x}^{2} }$

$\\ \\ \large\underline{ \underline{ \sf{ \red{solution:} }}}$

We know , x = 3 - 2√2

We will find (1/x).

$\\ \sf \: \frac{1}{x} = \frac{1}{3 - 2 \sqrt{2} } \\ \\ \\ \sf \: rationalising \: denominator \: .... \\ \sf \: rationalising \: factor \: is \: (3 + 2 \sqrt{2} \\ \\ \\ \sf \: \frac{1}{x} = \frac{1}{3 - 2 \sqrt{2} } \times \frac{3 + 2 \sqrt{2} }{3 + 2 \sqrt{2} } \\ \\ \\ \sf \: \frac{1}{x} = \frac{3 + 2 \sqrt{2} }{(3 - 2 \sqrt{2})(3 + 2 \sqrt{2} )} \\ \\ \\ \sf \: \frac{1}{x} = \frac{3 + 2 \sqrt{2} }{ {3}^{2} - ( {2 \sqrt{2}) }^{2} } \\ \\ \\ \sf \: \frac{1}{x} = \frac{3 + 2 \sqrt{2} }{9 - 8} \\ \\ \\ \sf \pink{ \frac{1}{x} = 3 + 2 \sqrt{2} }$

Now ,

$\\ \\ \sf \: x + \frac{1}{x} = 3 - 2 \sqrt{2} + 3 + 2 \sqrt{2} \\ \\ \\ \sf \: x + \frac{1}{x} = 6 \\ \\ \\ \sf \: squaring \: both \: sides...we \: get \\ \\ \\ \sf \: (x + { \frac{1}{x} )}^{2} = {6}^{2}$

We know ,

$\\ \boxed{ \bf \: {(a + b)}^{2} = {a}^{2} + {b}^{2} + 2ab }$

Here ,

• a = x
• b = 1/x

Putting values , we get ,

$\\ \\ \sf \: {x}^{2} + \frac{1}{ {x}^{2} } + 2( \cancel{x})( \frac{1}{ \cancel{x}} ) = 36 \\ \\ \\ \sf \: {x}^{2} + \frac{1}{ {x}^{2} } = 36 - 2 \\ \\ \\ \therefore\underline{\boxed{\sf \: \green{{x}^{2} + \frac{1}{ {x}^{2} } = 34}}}$

More identities :-

• (a-b)² = a² + b² - 2ab

• (a+b)(a-b) = a² - b²

• (a+x)(a+y) = a² + (x+y)a + xy