Rationalize the denominator of : \(\frac{\sqrt3+\sqrt2}{\sqrt3-\sqrt2}\)
1 Answer
Solution:
Here,
\(\frac{\sqrt3+\sqrt2}{\sqrt3-\sqrt2}\)
Multiplying both numerator and denominator by \(\sqrt3+\sqrt2\) , we get
\(\frac{\sqrt3+\sqrt2}{\sqrt3-\sqrt2}\times\frac{\sqrt3+\sqrt2}{\sqrt3+\sqrt2}\)
= \(\frac{{(\sqrt3+\sqrt2)}^2}{{(\sqrt3)}^2-{(\sqrt2)}^2}\)
= \(\frac{{(\sqrt3)}^2+2\times\sqrt3\times\sqrt2+{(\sqrt2)}^2}{3-2}\)
= \(\frac{3+2\times\sqrt6+2}{3-2}\)
= \(\frac{5+2\times\sqrt6}{1}\)
= \(5+2\sqrt6\)
Is the required answer
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