Question

Which of the following is equivalent to the radical expression below, when the denominator has been rationalized and X >_ 5?

Answer 1

Answer:

Option D is correct.

Step-by-step explanation:

$\frac{10}{\sqrt{x}-\sqrt{x-5}}$ We need to rationalize this term and find the answer.

To rationalize the term we multiply and divide the above expression by $\sqrt{x}+\sqrt{x-5}$

Solving:

$\frac{10}{\sqrt{x}-\sqrt{x-5}}\\=\frac{10}{\sqrt{x}-\sqrt{x-5}} * \frac{\sqrt{x}+\sqrt{x-5}}{\sqrt{x}+\sqrt{x-5}} \\Multiplying\\=\frac{10*(\sqrt{x}+\sqrt{x-5})}{\sqrt{x}-\sqrt{x-5}*\sqrt{x}+\sqrt{x-5}}\\=\frac{10*(\sqrt{x}+\sqrt{x-5})}{(\sqrt{x})^2-(\sqrt{x-5})^2}\\=\frac{10*(\sqrt{x}+\sqrt{x-5})}{x-(x-5)}\\=\frac{10*(\sqrt{x}+\sqrt{x-5})}{x-x+5)}\\=\frac{10*(\sqrt{x}+\sqrt{x-5})}{5}\\=2*(\sqrt{x}+\sqrt{x-5})$

So, Option D is correct.

Answer 2

Answer:

The correct option is D) $2\left(\sqrt{x}+\sqrt{x-5}\right)$.

Step-by-step explanation:

Consider the provided radical expression.

$\frac{10}{\sqrt{x}-\sqrt{x-5}}$

Multiply by the conjugate $\frac{\sqrt{x}+\sqrt{x-5}}{\sqrt{x}+\sqrt{x-5}}$

$\frac{10\left(\sqrt{x}+\sqrt{x-5}\right)}{\left(\sqrt{x}-\sqrt{x-5}\right)\left(\sqrt{x}+\sqrt{x-5}\right)}$

$\frac{10\left(\sqrt{x}+\sqrt{x-5}\right)}{\left(x-(x-5))}$

$\frac{10\left(\sqrt{x}+\sqrt{x-5}\right)}{5}$

$2\left(\sqrt{x}+\sqrt{x-5}\right)$

Hence, the correct option is D) $2\left(\sqrt{x}+\sqrt{x-5}\right)$.