Which of the following is equivalent to the radical expression below when the denominator has been rationalized and [tex][tex]$x \geq 5$[/tex][/tex]?

[tex]\frac{10}{\sqrt{x}-\sqrt{x-5}}[/tex]

A. [tex]2(\sqrt{x}+\sqrt{x-5})[/tex]

B. [tex]2(\sqrt{x}-\sqrt{x+5})[/tex]

C. [tex]2(\sqrt{x}-\sqrt{x-5})[/tex]

D. [tex]2(\sqrt{x}+\sqrt{x+5})[/tex]



Answer :

To rationalize the denominator of the expression [tex]\(\frac{10}{\sqrt{x} - \sqrt{x - 5}}\)[/tex] when [tex]\(x \geq 5\)[/tex], follow these steps:

1. Identify the denominator:
[tex]\(\sqrt{x} - \sqrt{x - 5}\)[/tex]

2. Multiply the numerator and denominator by the conjugate of the denominator:
The conjugate of [tex]\(\sqrt{x} - \sqrt{x - 5}\)[/tex] is [tex]\(\sqrt{x} + \sqrt{x - 5}\)[/tex].

Hence, we multiply the numerator and denominator by [tex]\(\sqrt{x} + \sqrt{x - 5}\)[/tex]:
[tex]\[ \frac{10}{\sqrt{x} - \sqrt{x - 5}} \cdot \frac{\sqrt{x} + \sqrt{x - 5}}{\sqrt{x} + \sqrt{x - 5}} \][/tex]

3. Simplify the expression:
The numerator becomes:
[tex]\[ 10 \cdot (\sqrt{x} + \sqrt{x - 5}) = 10 (\sqrt{x} + \sqrt{x - 5}) \][/tex]

The denominator becomes:
[tex]\[ (\sqrt{x} - \sqrt{x - 5})(\sqrt{x} + \sqrt{x - 5}) \][/tex]
Using the difference of squares formula, [tex]\(a^2 - b^2 = (a - b)(a + b)\)[/tex], where [tex]\(a = \sqrt{x}\)[/tex] and [tex]\(b = \sqrt{x - 5}\)[/tex], we get:
[tex]\[ (\sqrt{x})^2 - (\sqrt{x - 5})^2 = x - (x - 5) = x - x + 5 = 5 \][/tex]

Thus, the denominator simplifies to 5.

4. Final simplified form:
Now, the entire expression simplifies to:
[tex]\[ \frac{10 (\sqrt{x} + \sqrt{x - 5})}{5} = 2 (\sqrt{x} + \sqrt{x - 5}) \][/tex]

Therefore, the equivalent expression with a rationalized denominator is:
[tex]\[ \boxed{2(\sqrt{x} + \sqrt{x - 5})} \][/tex]

So the correct answer is:
A. [tex]\(2(\sqrt{x} + \sqrt{x - 5})\)[/tex]