Question

if x=3-2√2 find the value of x^4+1/x^4​

Answer 1

$\sf{x = 3 - 2 \sqrt{2} }$

$\sf \frac{1}{x} = \frac{1}{3 - 2 \sqrt{2} }$

rationalise the denominator

$\sf \frac{1}{x} = \frac{1}{3 - 2 \sqrt{2} } \times \frac{3 + 2 \sqrt{2} }{3 + 2 \sqrt{2} }$

$\sf \frac{1}{x} = \frac{3 + 2 \sqrt{2} }{ {3}^{2} -{ (2 \sqrt{2} )}^{2} }$

$\sf \frac{1}{x} = \frac{3 + 2 \sqrt{2} }{ 9 -8}$

$\sf \frac{1}{x} = {3 + 2 \sqrt{2} }$

$\sf {x + \frac{1}{x} = 3 - 2 \sqrt{2} + 3 + 2 \sqrt{2} }$

$\sf {x + \frac{1}{x} = 6}$

squaring on both sides

$\sf {{(x + \frac{1}{x})}^{2} = {6}^{2} }$

$\sf {{{x}^{2} + \frac{1}{{x}^{2} }} + 2= 36 }$

$\sf {{{x}^{2} + \frac{1}{{x}^{2} }} = 36 - 2}$

$\sf {{{x}^{2} + \frac{1}{{x}^{2} }} = 34}$

now again squaring on both sides

$\sf {{{({x}^{2} + \frac{1}{{x}^{2} }}) }^{2} = {34}^{2} }$

$\sf {{{x}^{4} + \frac{1}{{x}^{4} }} + 2 = 1156 }$

$\fbox{\sf {{{x}^{4} + \frac{1}{{x}^{4} }} = 1154 } }$