Question

30 POINTS!!! Which choice is equivalent to the fraction below when x is greater than or equal to 3?

Answer 1

Answer: C.

$\frac{9}{\sqrt{x} -\sqrt{x-3} }=\frac{9(\sqrt{x} +\sqrt{x-3} )}{x-(x-3)}=\frac{9(\sqrt{x} +\sqrt{x-3} )}{3}=3(\sqrt{x} +\sqrt{x-3})$

Step-by-step explanation:

Answer 2

The given fraction is equivalent to $3(\sqrt{x}+\sqrt{x-3})$. Then the correct answer is option C.

What is rationalization?

The elimination of radicals from an algebraic fraction's denominator is known as root rationalization in elementary algebra.

Rationalizing the given expression to get the solution.

$\dfrac{9}{\sqrt{x}-\sqrt{x-3}}=\dfrac{9}{\sqrt{x}-\sqrt{x-3}}\times \dfrac{\sqrt{x}+\sqrt{x-3}}{\sqrt{x}+\sqrt{x-3}}$

$\dfrac{9}{\sqrt{x}-\sqrt{x-3}}=\dfrac{9(\sqrt{x}-\sqrt{x-3})}{(\sqrt{x})^2-(\sqrt{x-3})^2}$

$\dfrac{9}{\sqrt{x}-\sqrt{x-3}}=\dfrac{9(\sqrt{x}-\sqrt{x-3})}{({x})-({x-3})}$

$\dfrac{9}{\sqrt{x}-\sqrt{x-3}}=\dfrac{9(\sqrt{x}-\sqrt{x-3})}{3}={3(\sqrt{x}-\sqrt{x-3}$

Hence, the correct answer is option C.

To know more about rationalization follow

https://brainly.com/question/14261303

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