Question
Simplify:$\frac{\sqrt5+\sqrt3}{\sqrt5-\sqrt3}+\frac{\sqrt5-\sqrt3}{\sqrt5+\sqrt3}$
✓
Answer
We know that rationalization factor for $\sqrt5-\sqrt3$ and $\sqrt5+\sqrt3$ are $\sqrt5+\sqrt3$ and $\sqrt5-\sqrt3$ respectively. We will multiply numerator and denominator of the given expression $\frac{\sqrt5+\sqrt3}{\sqrt5-\sqrt3}$ and $\frac{\sqrt5-\sqrt3}{\sqrt5+\sqrt3}$ by $\sqrt5+\sqrt3$ and $\sqrt5+\sqrt3$ respectively, to get$\frac{\sqrt5+\sqrt3}{\sqrt5-\sqrt3}\times\frac{\sqrt5+\sqrt3}{\sqrt5+\sqrt3}+\frac{\sqrt5-\sqrt3}{\sqrt5+\sqrt3}\times\frac{\sqrt5-\sqrt3}{\sqrt5-\sqrt3}\\=\frac{\big(\sqrt5\big)^2+\big(\sqrt3\big)^2+2\times\sqrt5\times\sqrt3}{\big(\sqrt5\big)^2-\big(\sqrt3\big)^2}+\frac{\big(\sqrt5\big)^2+\big(\sqrt3\big)^2-2\times\sqrt5\times\sqrt3}{\big(\sqrt5\big)^2-\big(\sqrt3\big)^2}$
$=\frac{5+3+2\sqrt{15}}{5-3}+\frac{5+3-2\sqrt{15}}{5-3}$
$=\frac{5+3+2\sqrt{15}+5+3-2\sqrt{15}}{2}$
$=\frac{16}{2}$
$=8$
Hence the given expression is simplified to 8.
$=\frac{5+3+2\sqrt{15}}{5-3}+\frac{5+3-2\sqrt{15}}{5-3}$
$=\frac{5+3+2\sqrt{15}+5+3-2\sqrt{15}}{2}$
$=\frac{16}{2}$
$=8$
Hence the given expression is simplified to 8.
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