Question

If a=3-2square root 2.Then find a^2 -1/a^2

Answer 1

Heya !!

$a = 3 - 2 \sqrt{2} \\ \\ \bf \: on \: squaring \: both \: the \: sides \: we \: get \\ \\ {(a)}^{2} = {(3 - 2 \sqrt{2}) }^{2} \\ \\ {a}^{2} = {(3)}^{2} + {(2 \sqrt{2} )}^{2} - 2(3)(2 \sqrt{2} ) \\ \\ {a}^{2} = 9 + 8 - 12 \sqrt{2} \\ \\ {a}^{2} = 17 - 12 \sqrt{2}$
$\frac{1}{a} = \frac{1}{3 - 2 \sqrt{2} }$

On rationalizing the denominator we get,

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

On squaring both the sides we get,

${( \frac{1}{a}) }^{2} = {(3 + 2 \sqrt{2}) }^{2} \\ \\ \frac{1}{ {a}^{2} } = {(3)}^{2} + {(2 \sqrt{2}) }^{2} + 2(3)(2 \sqrt{2} ) \\ \\ \frac{1}{ {a}^{2} } = 9 + 8 + 12 \sqrt{2} \\ \\ \frac{1}{ {a}^{2} } = 17 + 12 \sqrt{2}$

Now putting the values we get,

${a}^{2} - \frac{1}{ {a}^{2} } = (17 - 12 \sqrt{2} ) - (17 + 12 \sqrt{2} ) \\ \\ {a}^{2} - \frac{1}{ {a}^{2} } = 17 - 12 \sqrt{2} - 17 - 12 \sqrt{2} \\ \\ {a}^{2} - \frac{1}{ {a}^{2} } = - 24 \sqrt{2}$

Hope this helps ☺