Question

if x=6-√35find the value of x^2+1/x^2

Answer 1

Heya ✋

Let see your answer !!!!

Given that

$x = 6 - \sqrt{35} \\ x ^{2} + \frac{1}{x ^{2} } = what$
Solution

$\frac{1}{x} = \frac{1}{6 - \sqrt{35} } \\ = \frac{1 \times 6 + \sqrt{35} }{(6 - \sqrt{35)} \times (6 + \sqrt{35)} } \\ = \frac{6 + \sqrt{35} }{(6) ^{2} - ( \sqrt{35) ^{2} } } \\ = \frac{6 + \sqrt{35} }{36 - 35} \\ = 6 + \sqrt{35}$
$x ^{2} = (6 - \sqrt{35}) ^{2} \\ = (6) ^{2} + ( \sqrt{35) ^{2} } - 2 \times 6 \times \\ \sqrt{35} \\ = 36 + 35 - 12 \sqrt{35} \\ = 71 - 12 \sqrt{35}$
$\frac{1}{x ^{2} } = (6 + \sqrt{35}) ^{2} \\ = (6) ^{2} + ( \sqrt{35}) ^{2} + 2 \times 6 \times \\ \sqrt{35} \\ = 36 + 35 + 12 \sqrt{35} \\ = 71 + 12 \sqrt{35}$

Therefore ,

$x ^{2} + \frac{1}{x ^{2} } \\ = 71 - 12 \sqrt{35} + 71 + 12 \sqrt{35} \\ = 142$





Thanks :))))