Question

if a equals to 3 minus 2 root 2 find the value of a square minus 1 by a square

Answer 1

I hope this is helpful

Answer 2

Answer:

$The\:value \: of \: \\a^{2}-\frac{1}{a^{2}}\\=-24\sqrt{2}$

Step-by-step explanation:

$Given \: a = 3-2\sqrt{2}--(1)$

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

$ii) a+\frac{1}{a}\\=3-2\sqrt{2}+3+2\sqrt{2}\\=6---(3)$

$iii) a-\frac{1}{a}\\=3-2\sqrt{2}-(3+2\sqrt{2})\\=3-2\sqrt{2}-3-2\sqrt{2}\\=-4\sqrt{2}---(4)$

$Now,\\a^{2}-\frac{1}{a^{2}}\\=\left(a+\frac{1}{a}\right)\left(a-\frac{1}{a}\right)\\=6\times (-4\sqrt{2})\\=-24\sqrt{2}$

Therefore,

$The\:value \: of \: \\a^{2}-\frac{1}{a^{2}}\\=-24\sqrt{2}$

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