Which choice is equivalent to the fraction below when [tex]x[/tex] is an appropriate value?

Hint: Rationalize the denominator and simplify.

[tex]
\frac{4}{4-\sqrt{6x}}
[/tex]

A. [tex]\frac{8+2 \sqrt{6 x}}{16-6 x}[/tex]

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

C. [tex]\frac{8+2 \sqrt{6 x}}{8-3 x}[/tex]

D. [tex]\frac{8+2 \sqrt{6 x}}{8-6 x}[/tex]



Answer :

To solve the problem of finding the equivalent fraction for [tex]$ \frac{4}{4-\sqrt{6 x}} $[/tex] by rationalizing the denominator and simplifying, follow these steps:

### Step 1: Rationalize the Denominator
To rationalize the denominator, multiply both the numerator and the denominator by the conjugate of the denominator. The conjugate of [tex]\( 4 - \sqrt{6x} \)[/tex] is [tex]\( 4 + \sqrt{6x} \)[/tex].

Multiply the numerator and the denominator by [tex]\( 4 + \sqrt{6x} \)[/tex]:

[tex]\[ \frac{4}{4-\sqrt{6 x}} \cdot \frac{4+\sqrt{6 x}}{4+\sqrt{6 x}} = \frac{4(4+\sqrt{6 x})}{(4-\sqrt{6 x})(4+\sqrt{6 x})} \][/tex]

### Step 2: Simplify the Numerator
Expand the numerator:

[tex]\[ 4(4 + \sqrt{6x}) = 4 \cdot 4 + 4 \cdot \sqrt{6x} = 16 + 4\sqrt{6x} \][/tex]

### Step 3: Simplify the Denominator
Use the difference of squares formula for the denominator:

[tex]\[ (4 - \sqrt{6x})(4 + \sqrt{6x}) = 4^2 - (\sqrt{6x})^2 = 16 - 6x \][/tex]

### Step 4: Combine the Results
Substitute back the simplified numerator and denominator:

[tex]\[ \frac{16 + 4\sqrt{6x}}{16 - 6x} \][/tex]

### Step 5: Compare with Given Choices
Look at the given choices:

- A. [tex]\( \frac{8 + 2\sqrt{6x}}{16 - 6x} \)[/tex]
- B. [tex]\( \frac{2 + \sqrt{6x}}{4 - 6x} \)[/tex]
- C. [tex]\( \frac{8 + 2\sqrt{6x}}{8 - 3x} \)[/tex]
- D. [tex]\( \frac{8 + 2\sqrt{6x}}{8 - 6x} \)[/tex]

Simplifying and examination show that:
[tex]\[ 16 + 4\sqrt{6x} = 8 + 2(8 + 2\sqrt{6x}) \][/tex]

Combining the numerator and the constant:
[tex]\[ \frac{8 + 2\sqrt{6x}}{16 - 6x} \][/tex]

So the fraction equivalent to [tex]\( \frac{16 + 4 \sqrt{6x}}{16 - 6x} \)[/tex] matches choice A.

Thus, the answer is [tex]\( \boxed{A} \)[/tex].