Question

Which of the following is equivalent to the radical expression below, when the denominator has been rationalized and X >_ 6?

Answer 1

Answer:

  the correct choice is marked

Step-by-step explanation:

To rationalize the denominator, you multiply both numerator and denominator by the sum of the terms whose difference is the current denominator.

  $\dfrac{12}{\sqrt{x}-\sqrt{x-6}}=\dfrac{12(\sqrt{x}+\sqrt{x-6})}{(\sqrt{x}-\sqrt{x-6})(\sqrt{x}+\sqrt{x-6})}=\dfrac{12(\sqrt{x}+\sqrt{x-6})}{(\sqrt{x})^2-(\sqrt{x-6})^2}\\\\=\dfrac{12(\sqrt{x}+\sqrt{x-6})}{x-(x-6)}=\dfrac{12(\sqrt{x}+\sqrt{x-6})}{6}\\\\=\boxed{2(\sqrt{x}+\sqrt{x-6})}$