Question
If $\text{a}=3-2\sqrt{2},$ find the value of $\text{a}^2-\frac{1}{\text{a}^2}.$
✓
Answer
$\text{a}=3-2\sqrt{2}$$\Rightarrow\text{a}^2=\big(3-2\sqrt{2}\big)^2$
$=3^2-2\times3\times2\sqrt{2}+\big(2\sqrt{2}\big)^2$
$=9-12\sqrt{2}+8$
$=17-12\sqrt{2}$
$\Rightarrow\frac{1}{\text{a}^2}=\frac{1}{17-12\sqrt{2}}$
$=\frac{1}{17-12\sqrt{2}}\times\frac{17+12\sqrt{2}}{17+12\sqrt{2}}$
$=\frac{17+12\sqrt{2}}{17^2-\big(12\sqrt{2}\big)^2}$
$=\frac{17+12\sqrt{2}}{289-288}$
$=17+12\sqrt{2}$
$\Rightarrow\text{a}^2-\frac{1}{\text{a}^2}=\big(17-12\sqrt{2}\big)-\big(17+12\sqrt{2}\big)$
$=17-12\sqrt{2}-17-12\sqrt{2}$
$=-24\sqrt{2}$
$=3^2-2\times3\times2\sqrt{2}+\big(2\sqrt{2}\big)^2$
$=9-12\sqrt{2}+8$
$=17-12\sqrt{2}$
$\Rightarrow\frac{1}{\text{a}^2}=\frac{1}{17-12\sqrt{2}}$
$=\frac{1}{17-12\sqrt{2}}\times\frac{17+12\sqrt{2}}{17+12\sqrt{2}}$
$=\frac{17+12\sqrt{2}}{17^2-\big(12\sqrt{2}\big)^2}$
$=\frac{17+12\sqrt{2}}{289-288}$
$=17+12\sqrt{2}$
$\Rightarrow\text{a}^2-\frac{1}{\text{a}^2}=\big(17-12\sqrt{2}\big)-\big(17+12\sqrt{2}\big)$
$=17-12\sqrt{2}-17-12\sqrt{2}$
$=-24\sqrt{2}$
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