Question

If x=
$3 + 2 \sqrt{2}$
Find
$x {}^{2} + \frac{1}{x {}^{2} }$


​

Answer 1

Step-by-step explanation:

x=3+2root2

x^+1/x^2 =?

now 1/x = 1/3+2root2

after rationalisation

1/x = 3-2root2

now we know , (x+1/x)^2= x^2+1/x^2+2

so putting values

(3+2root2+3-2root2)^2=x^2+1/x^2+2

(6)^2=x^2+1/x^2+2

so, x^2+1/x^2 = 36-2

x^2+1/x^2= 34

your answer is 34

Answer 2

$\sf{given \: \: x = 3 + 2 \sqrt{2} }$

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

$\sf{rationalise \: the \: denominator} \\ \implies \sf{ \frac{1}{x} = \frac{1}{3 + 2 \sqrt{2} } \times \frac{3 - 2 \sqrt{2} }{3 - 2 \sqrt{2} } }$

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

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

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

$\sf{now}$

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

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

squaring on both sides

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

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

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

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

$\huge\star{\sf\small{Hope\:this\:helps\:you}}\huge\star$