Question

if a is equal to 9 + 4 under root 5 and b is equal to 1 upon a find a square + b square

Answer 1

Heya !!!

Identities used :

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



Given,

a = 9 + 4√5

b = 1/a

Putting the value of a we get,

$b = \frac{1}{9 + 4 \sqrt{5} }$

On rationalizing the denominator we get,

$b = \frac{1}{9 + 4 \sqrt{5} } \times \frac{9 - 4 \sqrt{5} }{9 - 4 \sqrt{5} } \\ \\ b = \frac{9 - 4 \sqrt{5} }{(9 + 4 \sqrt{5})(9 - 4 \sqrt{5} )} \\ \\ b = \frac{9 - 4 \sqrt{5} }{ {(9)}^{2} - {(4 \sqrt{5} )}^{2} } \\ \\ b = \frac{9 - 4 \sqrt{5} }{81 - 80} \\ \\ b = 9 - 4 \sqrt{5}$


Now,

${a}^{2} = {(9 + 4 \sqrt{5} )}^{2} \\ \\ {a}^{2} = {(9)}^{2} + {(4 \sqrt{5}) }^{2} + 2(9)(4 \sqrt{5} ) \\ \\ {a}^{2} = 81 + 80 + 72 \sqrt{5} \\ \\ {a}^{2} = 161 + 72 \sqrt{5}$

Also,

${b}^{2} = {(9 - 4 \sqrt{5}) }^{2} \\ \\ {b}^{2} = {(9)}^{2} + {(4 \sqrt{5}) }^{2} - 2(9)(4 \sqrt{5} ) \\ \\ {b}^{2} = 81 + 80 - 72 \sqrt{5} \\ \\ {b}^{2} = 161 - 72 \sqrt{5}$

Putting the values in (a)^2 + (b)^2


${a}^{2} + {b}^{2} = (161 + 72 \sqrt{5} ) + (161 - 72 \sqrt{5} ) \\ \\ {a}^{2} + {b}^{2} = 161 + 7 \sqrt{5} + 161 - 72 \sqrt{5} \\ \\ {a}^{2} + {b}^{2} = 161 + 161 \\ \\ {a}^{2} + {b}^{2} = 322$


Hope this helps ☺