Question

If x = 3+2√2 , find the value of x^2+1/x^2 ?

Answer 1

Hey friend,
Here is the answer you were looking for:
$x = 3 + 2 \sqrt{2} \\ \\ \frac{1}{x} = \frac{1}{3 + 2 \sqrt{2} } \\ \\ on \: rationalizing \: we \: get \\ \\ \frac{1}{x} = \frac{1}{3 + 2 \sqrt{2} } \times \frac{3 - 2 \sqrt{2} }{3 - 2 \sqrt{2} } \\ \\ using \: identity \\ (a + b)(a - b) = {a}^{2} - {b}^{2} \\ \\ = \frac{3 - 2 \sqrt{2} }{ {(3)}^{2} - {(2 \sqrt{2}) }^{2} } \\ \\ \frac{1}{x} = \frac{3 - 2 \sqrt{2} }{9 - 8} \\ \\ \frac{1}{x} = 3 - 2 \sqrt{2} \\ \\ x + \frac{1}{x} = (3 + 2 \sqrt{2} ) + (3 - 2 \sqrt{2} ) \\ \\ x + \frac{1}{x} = 3 + 2 \sqrt{2} + 3 - 2 \sqrt{2} \\ \\ x + \frac{1}{x} = 3 + 3 \\ \\ x + \frac{1}{x} = 6 \\ \\ squaring \: both \: side \\ \\ {(x + \frac{1}{x} )}^{2} = {(6)}^{2} \\ \\ {x}^{2} + \frac{1}{ {x}^{2} } + 2 \times x \times \frac{1}{x} = 36 \\ \\ {x}^{2} + \frac{1}{ {x}^{2} } + 2 = 36 \\ \\ {x}^{2} + \frac{1}{ {x}^{2} } = 36 - 2 \\ \\ {x}^{2} + \frac{1}{ {x}^{2} } = 34$

Hope this helps!!!

@Mahak24

Thanks...
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