Question
If $x=5-2 \sqrt{ 6} $, find $x^2+\frac{1}{x^2}$
✓
Answer
Given $x=5-2 \sqrt{6}$
We need to find $x^2+\frac{1}{x^2}$
Since $x=5-2 \sqrt{6}$, we have
$\frac{1}{x}=\frac{1}{5-2 \sqrt{6}}$
$ \Rightarrow \frac{1}{x}=\frac{1}{5-2 \sqrt{6}} \times \frac{5+2 \sqrt{6}}{5+2 \sqrt{6}}$
$ \Rightarrow \frac{1}{x}=\frac{5-2 \sqrt{6}}{(5-2 \sqrt{6})(5+2 \sqrt{6})}$
$ \Rightarrow \frac{1}{x}=\frac{5+2 \sqrt{6}}{5^2-(2 \sqrt{6})^2}$
$ \Rightarrow \frac{1}{x}=\frac{5+2 \sqrt{6}}{25-24}$
$ \Rightarrow \frac{1}{x}=\frac{5+2 \sqrt{6}}{1}$
$\Rightarrow \frac{1}{x}=(5+2 \sqrt{6}) \dots....(1)$
Thus, $\left(x-\frac{1}{x}\right)=(5- \not 2 \sqrt{\not6})-(5+\not 2 \sqrt{\not 6})=10$
$\left(x^2\right)+\left(\frac{1}{x^2}\right)=(-4 \sqrt{6})^2+2$
$ \Rightarrow\left(x^2\right)+\left(\frac{1}{x^2}\right)=96+2$
$ \Rightarrow\left(x^2\right)+\left(\frac{1}{x^2}\right)=98$
We need to find $x^2+\frac{1}{x^2}$
Since $x=5-2 \sqrt{6}$, we have
$\frac{1}{x}=\frac{1}{5-2 \sqrt{6}}$
$ \Rightarrow \frac{1}{x}=\frac{1}{5-2 \sqrt{6}} \times \frac{5+2 \sqrt{6}}{5+2 \sqrt{6}}$
$ \Rightarrow \frac{1}{x}=\frac{5-2 \sqrt{6}}{(5-2 \sqrt{6})(5+2 \sqrt{6})}$
$ \Rightarrow \frac{1}{x}=\frac{5+2 \sqrt{6}}{5^2-(2 \sqrt{6})^2}$
$ \Rightarrow \frac{1}{x}=\frac{5+2 \sqrt{6}}{25-24}$
$ \Rightarrow \frac{1}{x}=\frac{5+2 \sqrt{6}}{1}$
$\Rightarrow \frac{1}{x}=(5+2 \sqrt{6}) \dots....(1)$
Thus, $\left(x-\frac{1}{x}\right)=(5- \not 2 \sqrt{\not6})-(5+\not 2 \sqrt{\not 6})=10$
$\left(x^2\right)+\left(\frac{1}{x^2}\right)=(-4 \sqrt{6})^2+2$
$ \Rightarrow\left(x^2\right)+\left(\frac{1}{x^2}\right)=96+2$
$ \Rightarrow\left(x^2\right)+\left(\frac{1}{x^2}\right)=98$
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