Question
Find rational numbers a and b such that:
$\frac{2-\sqrt{5}}{2+\sqrt{5}}=\text{a}\sqrt{5}+\text{b}$
$\frac{2-\sqrt{5}}{2+\sqrt{5}}=\text{a}\sqrt{5}+\text{b}$
✓
Answer
$\frac{2-\sqrt{5}}{2+\sqrt{5}}=\text{a}\sqrt{5}+\text{b}$
we have,
$\frac{2-\sqrt{5}}{2+\sqrt{5}}$
$=\frac{2-\sqrt{5}}{2+\sqrt{5}}\times\frac{2-\sqrt{5}}{2-\sqrt{5}}$
$=\frac{\big(2-\sqrt{5}\big)^2}{(2)^2-\big(\sqrt{5}\big)^2}$
$=\frac{(2)^2-2\times2\sqrt{5}+\big(\sqrt{5}\big)^2}{4-5}$
$=\frac{4-4\sqrt{5}+5}{-1}$
$=-\big(-4\sqrt{5}+9\big)$
$=4\sqrt{5}-9$
Now,
$\frac{2-\sqrt{5}}{2+\sqrt{5}}=\text{a}\sqrt{5}+\text{b}$
$\Rightarrow4\sqrt{5}-9=\text{a}\sqrt{5}+\text{b}$
$\Rightarrow\text{a}=4$ and $\text{b}=-9$
we have,
$\frac{2-\sqrt{5}}{2+\sqrt{5}}$
$=\frac{2-\sqrt{5}}{2+\sqrt{5}}\times\frac{2-\sqrt{5}}{2-\sqrt{5}}$
$=\frac{\big(2-\sqrt{5}\big)^2}{(2)^2-\big(\sqrt{5}\big)^2}$
$=\frac{(2)^2-2\times2\sqrt{5}+\big(\sqrt{5}\big)^2}{4-5}$
$=\frac{4-4\sqrt{5}+5}{-1}$
$=-\big(-4\sqrt{5}+9\big)$
$=4\sqrt{5}-9$
Now,
$\frac{2-\sqrt{5}}{2+\sqrt{5}}=\text{a}\sqrt{5}+\text{b}$
$\Rightarrow4\sqrt{5}-9=\text{a}\sqrt{5}+\text{b}$
$\Rightarrow\text{a}=4$ and $\text{b}=-9$
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