Question

if x is equal to 3 + root 5 upon 2 then find X square + 1 by x square

Answer 1

I hope this answer would be helpful for you

Answer 2

$i ) x = \frac{(3+\sqrt{5}}{2} \: --(1)$

$ii ) \frac{1}{x} = \frac{2}{(3+\sqrt{5})}$

$= \frac{2(3-\sqrt{5})}{(3+\sqrt{5})(3-\sqrt{5})}$

$= \frac{2(3-\sqrt{5})}{3^{2}-(\sqrt{5})^{2}}$

$= \frac{2(3-\sqrt{5})}{9- 5}$

$= \frac{2(3-\sqrt{5})}{4}$

$= \frac{(3-\sqrt{5})}{2} \: ---(2)$

$iii ) x + \frac{1}{x} \\= \frac{(3+\sqrt{5})}{2} + \frac{(3-\sqrt{5})}{2} \\= \frac{3+\sqrt{5} + 3-\sqrt{5}}{2} \\= \frac{6}{2} = 3\: ---(3)$

$iv ) Now , the \: value \:of \: x^{2} + \frac{1}{x^{2}} \\= \Big( x + \frac{1}{x} \Big)^{2} - 2 \times x \times \frac{1}{x} \\= 3^{2} - 2\\= 9 - 2 \\= 7$

Therefore.,

$\red { The \: value \:of \: x^{2} + \frac{1}{x^{2}} } \green {= 7 }$

•••♪