Rationalize the denominator of (sqrt(5)-sqrt(7))/(sqrt(3)-sqrt(2))

Question

No matter how many tutorials I watch, I can't understand how to rationalise a fraction.

Using

${\sqrt{5}-\sqrt{7}}\over {\sqrt{3}-\sqrt{2}}$

as an example, please explain step by step how I would rationalise the denominator.

Answer 1

For all numbers $a$ and $b$, we have: $$(a+b)\cdot(a-b)=a^2-b^2$$ Also a number by divided by itself is $1$ and a number multiplied by $1$ is itself. Now: $$\dfrac{\sqrt{5}-\sqrt{7}}{\sqrt{3}-\sqrt{2}}=\dfrac{\sqrt{5}-\sqrt{7}}{\sqrt{3}-\sqrt{2}}\cdot 1=\dfrac{\sqrt{5}-\sqrt{7}}{\sqrt{3}-\sqrt{2}}\cdot\dfrac{\sqrt{3}+\sqrt{2}}{\sqrt{3}+\sqrt{2}}=\dfrac{(\sqrt{5}-\sqrt{7})(\sqrt{3}+\sqrt{2})}{(\sqrt{3}-\sqrt{2})(\sqrt{3}+\sqrt{2})}=\dfrac{\sqrt{15}+\sqrt{10}-\sqrt{21}-\sqrt{14}}{\sqrt{3}^2+\sqrt{2}^2}=\dfrac{\sqrt{15}+\sqrt{10}-\sqrt{21}-\sqrt{14}}{5}$$

Answer 2

Multiply numerator and denominator by $\sqrt3+\sqrt2$