Question 9 of 10

Which choice is equivalent to the fraction below when [tex]$x$[/tex] is an appropriate value? Hint: Rationalize the denominator and simplify.

[tex]\frac{2}{2-\sqrt{6 x}}[/tex]

A. [tex]\frac{2+\sqrt{6 x}}{2-3 x}[/tex]

B. [tex]\frac{2+\sqrt{6 x}}{4-3 x}[/tex]

C. [tex]\frac{2+\sqrt{6 x}}{2-6 x}[/tex]

D. [tex]\frac{2+\sqrt{6 x}}{4-6 x}[/tex]



Answer :

To determine which choice is equivalent to the given fraction [tex]\(\frac{2}{2 - \sqrt{6x}}\)[/tex], we need to rationalize the denominator and simplify it.

The expression given is:
[tex]\[ \frac{2}{2 - \sqrt{6x}} \][/tex]

To rationalize the denominator, we will multiply both the numerator and the denominator by the conjugate of the denominator. The conjugate of [tex]\(2 - \sqrt{6x}\)[/tex] is [tex]\(2 + \sqrt{6x}\)[/tex]. Multiplying by this conjugate will help to eliminate the square root in the denominator.

So, we multiply the fraction by [tex]\(\frac{2 + \sqrt{6x}}{2 + \sqrt{6x}}\)[/tex]:
[tex]\[ \frac{2}{2 - \sqrt{6x}} \cdot \frac{2 + \sqrt{6x}}{2 + \sqrt{6x}} = \frac{2(2 + \sqrt{6x})}{(2 - \sqrt{6x})(2 + \sqrt{6x})} \][/tex]

First, let's simplify the numerator:
[tex]\[ 2(2 + \sqrt{6x}) = 4 + 2\sqrt{6x} \][/tex]

Next, we simplify the denominator. The denominator is a difference of squares:
[tex]\[ (2 - \sqrt{6x})(2 + \sqrt{6x}) = 2^2 - (\sqrt{6x})^2 = 4 - 6x \][/tex]

Now, putting it all together, we have:
[tex]\[ \frac{4 + 2\sqrt{6x}}{4 - 6x} \][/tex]

Thus, the rationalized and simplified form of the given fraction [tex]\(\frac{2}{2 - \sqrt{6x}}\)[/tex] is:
[tex]\[ \frac{4 + 2\sqrt{6x}}{4 - 6x} \][/tex]

To match one of the answer choices, we can simplify by dividing both the numerator and the denominator by 2:
[tex]\[ \frac{4 + 2\sqrt{6x}}{4 - 6x} = \frac{2(2 + \sqrt{6x})}{2(2 - 3x)} = \frac{2 + \sqrt{6x}}{2 - 3x} \][/tex]

Therefore, the correct choice is:
[tex]\[ \boxed{\frac{2 + \sqrt{6x}}{4 - 6x}} \][/tex]

So, the correct answer is:
D. [tex]\(\frac{2 + \sqrt{6x}}{4 - 6x}\)[/tex]