Question

If x=3x+2 √2 find the value of x^2+1/x^2
Please give detailed solutions

Answer 1

Solution

Given

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

To Find

$x^{2}\:+\:\dfrac{1}{x^{2}}$

Steps to follow

$\dfrac{1}{x}\: = \: \dfrac{1}{3\: + \: 2\sqrt{2}}$

$\dfrac{1}{x}\: = \: \dfrac{1}{3\: + \: 2\sqrt{2}}\times\dfrac{3\: - \: 2\sqrt{2}}{3\: - \: 2\sqrt{2}}$

$\dfrac{1}{x} \: = \: \dfrac{3\: - 2\sqrt{2}}{3^{2} \: - \: (2\sqrt{2})^{2}}$

$\dfrac{1}{x}\: = \: \dfrac{3\: - 2\sqrt{2}}{9\: - \: 4\times 2}$

$\dfrac{1}{x}\: = \: \dfrac{3\: - 2\sqrt{2}}{9\: - \: 8}$

$\dfrac{1}{x}\: = \: \dfrac{3\: - 2\sqrt{2}}{1}$

$\dfrac{1}{x}\: = \: 3\: - 2\sqrt{2}$

$x^{2}\:+\:(\dfrac{1}{x})^{2}$

$\implies (3\: + \: 2\sqrt{2})^{2} \: + \: (3\: - \: 2\sqrt{2})^{2}$

$\implies 9 \: + \: 8 \: + \: 12\sqrt{2} \: + \: 9 \: + \: 8 \: - \: 12\sqrt{2}$

eliminate 12√2

$\implies 9 \: + \: 8 \: + \:+ \: 9 \: + \: 8$

$\implies 18 \: + \: 16$

$\implies 34$

$\boxed{x^{2}\:+\:\dfrac{1}{x^{2}}\: =\: 34}$

Answer 2

Given: $x=3+2\sqrt{2}$

To find: $x^{2}+\frac{1}{x^{2}}$

Solution 1.

Here, $x=3+2\sqrt{2}$

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

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

$\quad\quad=\frac{3-2\sqrt{2}}{9-8}$

$\quad\quad=\frac{3-2\sqrt{2}}{1}$

$\quad\quad=3-2\sqrt{2}$

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

$\quad=(3+2\sqrt{2})^{2}+(3-2\sqrt{2})^{2}$

$\quad=9+12\sqrt{2}+8+9-12\sqrt{2}+8$

$\quad=34$

$\Rightarrow \boxed{\color{blue}x^{2}+\frac{1}{x^{2}}=34}$

Solution 2.

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

$\quad=\left(x+\frac{1}{x}\right)^{2}-2*x*\frac{1}{x}$

$\quad=(3+2\sqrt{2}+3-2\sqrt{2})^{2}-2$

$\quad=6^{2}-2$

$\quad=36-2$

$\quad=34$

$\Rightarrow \boxed{\color{blue}x^{2}+\frac{1}{x^{2}}=34}$