Question

If x =
$3 + \sqrt{8}$
Find
${x}^{2} + \frac{1}{ {x}^{2} }$
​

Answer 1

x= 3+√8

1/x=3-√8.

(x+1/x)^2= x^2+1/x√2 +2

3+√8+3-√8=x^2+1/x^2+2

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

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

34=x^2+1/x^2.

so, 34 is answer.

Answer 2

$\mathfrak{\large{\underline{\underline{Given:-}}}}$

$3 + \sqrt{8}$

$\mathfrak{\large{\underline{\underline{To find:-}}}}$

${x}^{2} + \frac{1}{ {x}^{2} }$

$\mathfrak{\large{\underline{\underline{solution:-}}}}$

x = 3 + √8 , $\frac{1}{x} = 3 - \sqrt{8}$

$\implies$ $\bold{(x + \frac{1}{x} )^{2} = {x}^{2} + \frac{1}{ {x}^{2} } + 2x. \frac{1}{x} }$

$\implies$ $\bold{ (3 + \sqrt{8} + 3 - \sqrt{8} ) ^{2} = {x}^{2} + \frac{1}{ {x}^{2} } + 2}$

$\implies$ $\bold{ {6}^{2} - 2 = {x}^{2} + \frac{1}{ {x}^{2} } }$

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

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