Answer: Choice C) [tex]\frac{8+2\sqrt{6x}}{8-3x}[/tex]
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Work Shown:
[tex]y = \frac{4}{4-\sqrt{6x}}\\\\y = \frac{4(4+\sqrt{6x})}{(4-\sqrt{6x})(4+\sqrt{6x})}\\\\y = \frac{4(4+\sqrt{6x})}{4^2 - (\sqrt{6x})^2}\\\\y = \frac{4(4+\sqrt{6x})}{16-6x}\\\\y = \frac{2*2(4+\sqrt{6x})}{2(8-3x)}\\\\y = \frac{2(4+\sqrt{6x})}{8-3x}\\\\y = \frac{8+2\sqrt{6x}}{8-3x}\\\\[/tex]
This shows why choice C is the answer.
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Notes:
- If you have a+sqrt(b) in the denominator, multiply top and bottom by a-sqrt(b) which is the conjugate, and that will rationalize the denominator.
- In the second step, I multiplied top and bottom by 4+sqrt(6x) to rationalize the denominator
- In step 3, I used the difference of squares rule. In the step afterward, the square root is eliminated.
