Question

If x is equal to 3 minus 2 root 2 find X square + 1 by x square

Answer 1

Heya !!

Identity used :

$(x + y)(x - y) = {x}^{2} - {y}^{2}$

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

On rationalizing the denominator we get,


$\frac{1}{x} = \frac{1}{3 - 2 \sqrt{2} } \times \frac{3 + 2 \sqrt{2} }{3 + 2 \sqrt{2} } \\ \\ \frac{1}{x} = \frac{3 + 2 \sqrt{2} }{ {(3)}^{2} - {(2 \sqrt{2}) }^{2} } \\ \\ \frac{1}{x} = \frac{3 + 2 \sqrt{2} }{9 - 8} \\ \\ \frac{1}{x} = 3 + 2 \sqrt{2} \\ \\ x + \frac{1}{x} = (3 - 2 \sqrt{2} ) + (3 + 2 \sqrt{2} ) \\ \\ x + \frac{1}{x} = 3 - 2 \sqrt{2} + 3 + 2 \sqrt{2} \\ \\ x + \frac{1}{x} = 6$

On squaring both the sides we get,

${(x + \frac{1}{x}) }^{2} = {(6)}^{2} \\ \\ {(x)}^{2} + {( \frac{1}{x}) }^{2} + 2 \times x \times \frac{1}{x} = 36 \\ \\ {x}^{2} + \frac{1}{ {x}^{2} } + 2 = 36 \\ \\ {x}^{2} + \frac{1}{ {x}^{2} } = 36 - 4 \\ \\ {x}^{2} + \frac{1}{ {x}^{2} } = 34$

Hope this helps ☺

Answer 2

hope this helps you