Question
If $\text{x}=9-4\sqrt{5},$ find the value of $\text{x}^2-\frac{1}{\text{x}^2}.$
✓
Answer
$\text{x}=9-4\sqrt{5}$
$\Rightarrow\frac{1}{\text{x}}=\frac{1}{9-4\sqrt{5}}=\frac{1}{9-4\sqrt{5}}\times\frac{9+4\sqrt{5}}{9+4\sqrt{5}}\\=\frac{9+4\sqrt{5}}{9^2-\big(4\sqrt{5}\big)^2}=\frac{9+4\sqrt{5}}{81-80}=9+4\sqrt{5}$
$\Rightarrow\text{x}+\frac{1}{\text{x}}=9-4\sqrt{5}+9+4\sqrt{5}=18$
$\Rightarrow\Big(\text{x}+\frac{1}{\text{x}}\Big)^2=18^2=324$
$\Rightarrow\Big(\text{x}^2+\frac{1}{\text{x}^2}\Big)^2+2\times\text{x}\times\frac{1}{\text{x}}=324$
$\Rightarrow\Big(\text{x}^2+\frac{1}{\text{x}^2}\Big)+2=324$
$\Rightarrow\Big(\text{x}^2+\frac{1}{\text{x}^2}\Big)=322$
$\Rightarrow\frac{1}{\text{x}}=\frac{1}{9-4\sqrt{5}}=\frac{1}{9-4\sqrt{5}}\times\frac{9+4\sqrt{5}}{9+4\sqrt{5}}\\=\frac{9+4\sqrt{5}}{9^2-\big(4\sqrt{5}\big)^2}=\frac{9+4\sqrt{5}}{81-80}=9+4\sqrt{5}$
$\Rightarrow\text{x}+\frac{1}{\text{x}}=9-4\sqrt{5}+9+4\sqrt{5}=18$
$\Rightarrow\Big(\text{x}+\frac{1}{\text{x}}\Big)^2=18^2=324$
$\Rightarrow\Big(\text{x}^2+\frac{1}{\text{x}^2}\Big)^2+2\times\text{x}\times\frac{1}{\text{x}}=324$
$\Rightarrow\Big(\text{x}^2+\frac{1}{\text{x}^2}\Big)+2=324$
$\Rightarrow\Big(\text{x}^2+\frac{1}{\text{x}^2}\Big)=322$
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