Question
If $\text{x}=\frac{5-\sqrt{21}}{2},$ find the value of $\text{x}+\frac{1}{\text{x}}.$
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Answer
$\text{x}=\frac{5-\sqrt{21}}{2}$
$\Rightarrow\frac{1}{\text{x}}=\frac{2}{5-\sqrt{21}}$
$=\frac{2}{5-\sqrt{21}}\times\frac{5+\sqrt{21}}{5+\sqrt{21}}$
$=\frac{2\big(5+\sqrt{21}\big)}{5^2-\big(\sqrt{21}\big)^2}$
$=\frac{2\big(5+\sqrt{21}\big)}{25-21}$
$=\frac{2\big(5+\sqrt{21}\big)}{4}$
$=\frac{5+\sqrt{21}}{2}$
$\Rightarrow\text{x}+\frac{1}{\text{x}}=\frac{5-\sqrt{21}}{2}+\frac{5+\sqrt{21}}{2}$
$=\frac{5-\sqrt{21}+5-\sqrt{21}}{2}$
$=\frac{10}{2}$
$=5$
$\Rightarrow\frac{1}{\text{x}}=\frac{2}{5-\sqrt{21}}$
$=\frac{2}{5-\sqrt{21}}\times\frac{5+\sqrt{21}}{5+\sqrt{21}}$
$=\frac{2\big(5+\sqrt{21}\big)}{5^2-\big(\sqrt{21}\big)^2}$
$=\frac{2\big(5+\sqrt{21}\big)}{25-21}$
$=\frac{2\big(5+\sqrt{21}\big)}{4}$
$=\frac{5+\sqrt{21}}{2}$
$\Rightarrow\text{x}+\frac{1}{\text{x}}=\frac{5-\sqrt{21}}{2}+\frac{5+\sqrt{21}}{2}$
$=\frac{5-\sqrt{21}+5-\sqrt{21}}{2}$
$=\frac{10}{2}$
$=5$
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