Question

if a is equal to 3 + under root 8 find the value of A square + 1 upon S square​

Answer 1

Answer:

$\sf a = 3 + \sqrt{8} \\ \\ \sf a = 3 + 2 \sqrt{2} \\ \\ \sf \frac{1}{a} = \frac{1}{3 + 2 \sqrt{2} } \times \frac{3 - 2 \sqrt{2} }{3 - 2 \sqrt{2} } \\ \\ \sf \frac{1 }{a} = \frac{3 - 2 \sqrt{2} }{(3) {}^{2} - 2 \sqrt{2}) {}^{2} } \\ \\ \sf \frac{1}{a} = 3 - 2 \sqrt{2} \\ \\ \sf a + \frac{1}{a} = 3 + 2 \sqrt{2} + 3 - 2 \sqrt{2} \\ \\ \sf a + \frac{1}{a} = 3 + 3 \\ \\ \sf a + \frac{1}{a} = 6 \\ \\ \bf squaring \: both \: sides - \\ \\ \sf ( a + \frac{1}{a} ) {}^{2} = (6) {}^{2} \\ \\ \sf a {}^{2} + \frac{1}{a {}^{2} } + 2 = 36 \\ \\ \sf a {}^{2} + \frac{1}{a {}^{2} } = 36 - 2 \\ \\ \boxed{ \red{ \tt a {}^{2} + \frac{1}{a {}^{2} } = 34}}$

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