Question

if a is equal to 8 + 3 root 7 and b is equal to 1 / find the value of a square + b square

Answer 1

Heya !!!

Identities used :

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


$a = 8 + 3 \sqrt{7}$

On squaring both the sides we get,

${a}^{2} = {(8 + 3 \sqrt{7} )}^{2} \\ \\ {a}^{2} = {(8)}^{2} + {(3 \sqrt{7} )}^{2} + 2(8)(3 \sqrt{7} ) \\ \\ {a}^{2} = 64 + 63 + 48 \sqrt{7} \\ \\ {a}^{2} = 127 + 48 \sqrt{7}$

$b = \frac{1}{a} \\ \\ b = \frac{1}{8 + 3 \sqrt{7} }$

On rationalizing the denominator we get,

$b = \frac{1}{8 + 3 \sqrt{7} } \times \frac{8 - 3 \sqrt{7} }{8 - 3 \sqrt{7} } \\ \\ b = \frac{8 - 3 \sqrt{7} }{ {(8)}^{2} - {(3 \sqrt{7} )}^{2} } \\ \\ b = \frac{8 - 3 \sqrt{7} }{64 - 63} \\ \\ b = 8 - 3 \sqrt{7}$

On squaring both the sides we get,

${b}^{2} = {(8 - 3 \sqrt{7}) }^{2} \\ \\ {b}^{2} = {(8)}^{2} + {(3 \sqrt{7}) }^{2} - 2(8)(3 \sqrt{7} ) \\ \\ {b}^{2} = 64 + 63 - 48 \sqrt{7} \\ \\ {b}^{2} = 127 - 48 \sqrt{7}$

${a}^{2} + {b}^{2} \\ \\ = (127 + 48 \sqrt{7} ) + (127 - 48 \sqrt{7} ) \\ \\ = 127 + 48 \sqrt{7} + 127 - 48 \sqrt{7} \\ \\ = 127 + 127 \\ \\ = 254$

Hope this helps ☺