Question
Find rational numbers $a$ and $b$ such that: $\frac{\sqrt{3}+\sqrt{2}}{\sqrt{3}-\sqrt{2}}=\text{a}+\text{b}\sqrt{6}$
✓
Answer
$\frac{\sqrt{3}+\sqrt{2}}{\sqrt{3}-\sqrt{2}}=\text{a}+\text{b}\sqrt{6}$
we have, $\frac{\sqrt{3}+\sqrt{2}}{\sqrt{3}-\sqrt{2}}$
$=\frac{\sqrt{3}+\sqrt{2}}{\sqrt{3}-\sqrt{2}}\times\frac{\sqrt{3}+\sqrt{2}}{\sqrt{3}+\sqrt{2}}$
$=\frac{\big(\sqrt{3}+\sqrt{2}\big)^2}{\big(\sqrt{3}\big)^2-\big(\sqrt{2}\big)^2}$
$=\frac{\big(\sqrt{3}\big)^2+2\times\sqrt{2}\times\sqrt{3}+\big(\sqrt{2}\big)^2}{3-2}$
$=\frac{3+2\sqrt{6}+2}{1}$
$=5+2\sqrt{6}$
Now, $\frac{\sqrt{3}+\sqrt{2}}{\sqrt{3}-\sqrt{2}}=\text{a}+\text{b}\sqrt{6}$
$\Rightarrow\text{a}=5$ and $\text{b}=2$
we have, $\frac{\sqrt{3}+\sqrt{2}}{\sqrt{3}-\sqrt{2}}$
$=\frac{\sqrt{3}+\sqrt{2}}{\sqrt{3}-\sqrt{2}}\times\frac{\sqrt{3}+\sqrt{2}}{\sqrt{3}+\sqrt{2}}$
$=\frac{\big(\sqrt{3}+\sqrt{2}\big)^2}{\big(\sqrt{3}\big)^2-\big(\sqrt{2}\big)^2}$
$=\frac{\big(\sqrt{3}\big)^2+2\times\sqrt{2}\times\sqrt{3}+\big(\sqrt{2}\big)^2}{3-2}$
$=\frac{3+2\sqrt{6}+2}{1}$
$=5+2\sqrt{6}$
Now, $\frac{\sqrt{3}+\sqrt{2}}{\sqrt{3}-\sqrt{2}}=\text{a}+\text{b}\sqrt{6}$
$\Rightarrow\text{a}=5$ and $\text{b}=2$
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