Question

RATIONALISE THE DENOMINATOR ROOT 3+ROOT 2/ROOT3-ROOT2

Answer 1

Answer:

$\frac{\sqrt{3}+\sqrt{2}}{\sqrt{3}-\sqrt{2}}=5+2\sqrt{6}$

Step-by-step explanation:

$Given \: \frac{\sqrt{3}+\sqrt{2}}{\sqrt{3}-\sqrt{2}}$

Multiply numerator and denominator by (√3+√2), we get

$=\frac{(\sqrt{3}+\sqrt{2})(\sqrt{3}+\sqrt{2})}{(\sqrt{3}-\sqrt{2})(\sqrt{3}+\sqrt{2})}$

$=\frac{\big(\sqrt{3}+\sqrt{2}\big)^{2}}{(\sqrt{3})^{2}-(\sqrt{2})^{2}}$

$=\frac{(\sqrt{3})^{2}+(\sqrt{2})^{2}+2\times \sqrt{3}\times \sqrt{2}}{3-2}$

$=\frac{3+2+2\sqrt{6}}{1}$

$=5+2\sqrt{6}$

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