Question

If a=8+3 root 7,find a square + 1 by a square

Answer 1

Answer:

$256$

Step-by-step explanation:

$a = 8 + 3 \sqrt{7} \\ \frac{1}{a } = \frac{1}{8 + 3 \sqrt{7} } \\ = \frac{1}{8 + 3 \sqrt{7}} \times \frac{8 - 3 \sqrt{7} }{8 - 3 \sqrt{7} } \\ = \frac{8 - 3 \sqrt{7} }{(8) {}^{2} -( 3 \sqrt{7} ) {}^{2} } \\ = \frac{8 - 3 \sqrt{7}}{64 - 63 } \\ = 8 - 3 \sqrt{7} \\ thus - \\ a + \frac{1}{a} = 8 + 3 \sqrt{7 }+ 8 - 3 \sqrt{7} = 16 \\ therefore - \\ (a + \frac{1}{a} ) {}^{2} = (16) {}^{2} = 256$