Can anyone help me understand the simplification of $\frac{\sqrt 3 + \sqrt 2}{\sqrt 3 - \sqrt 2}\;$?

Question

Can anyone help me understand the following simplification of the fraction? $$\dfrac{\sqrt 3 + \sqrt 2}{\sqrt 3 - \sqrt 2} = 5 + 2\sqrt 6$$

I can't understand how to simplify the left-hand side to get the right-hand side. Your help would be great.

Thanks in advance.

Answer 1

Multiply numerator and denominator by $(\sqrt 3+ \sqrt 2)$

You'll have a difference of squares in the denominator: $$(a - b)(a+b) = a^2 - b^2$$

$$\frac{\sqrt{3}+\sqrt{2}}{\sqrt{3}-\sqrt{2}}\cdot \frac{\sqrt 3 + \sqrt 2}{\sqrt 3 + \sqrt 2} = \frac{(\sqrt 3 + \sqrt 2)^2}{3 - 2} = \;3 + 2\sqrt{3\cdot 2} + 2 = 5 +2\sqrt 6$$

Answer 2

Multiply the fraction by $\frac{\sqrt{3}+\sqrt{2}}{\sqrt{3}+\sqrt{2}}$ and use $a^2-b^2 = (a-b)(a+b)$. So $$\frac{\sqrt{3}+\sqrt{2}}{\sqrt{3}-\sqrt{2}} =\frac{\sqrt{3}+\sqrt{2}}{\sqrt{3}-\sqrt{2}}\frac{\sqrt{3}+\sqrt{2}}{\sqrt{3}+\sqrt{2}} = \frac{(\sqrt{3}+\sqrt{2})^2}{1} = 3 + 2\sqrt{3}\sqrt{2}+2 = 5 + 2\sqrt{6}$$

Answer 3

So if you consider:: $$\frac{\sqrt{3} + \sqrt{2}}{\sqrt{3} - \sqrt{2}} \times \frac{\sqrt{3} + \sqrt{2}}{\sqrt{3} + \sqrt{2}} = \frac{(\sqrt{3} + \sqrt{2})^2}{(\sqrt{3} + \sqrt{2})(\sqrt{3} - \sqrt{2})} = \frac{3+2\sqrt{3}\sqrt{2}+2}{3-2} = 5+2\sqrt{6}.$$