Answer:
A.) 2√6 + 4
Step-by-step explanation:
The reason for the direction to use a conjugate to rationalize the denominator is that multiplying by the conjugate lets you take advantage of the factoring of the difference of squares:
(a +b)(a -b) = a² -b²
When "a" or "b" involves a square root, this will eliminate the radical.
So, you have ...
[tex]\dfrac{2\sqrt{2}}{\sqrt{3}-\sqrt{2}}=\dfrac{2\sqrt{2}(\sqrt{3}+\sqrt{2})}{(\sqrt{3}-\sqrt{2})(\sqrt{3}+\sqrt{2})}\\\\=\dfrac{2\sqrt{2\cdot 3}+2\sqrt{2\cdot 2}}{3-2}=2\sqrt{6}+4[/tex]
