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Which choice is equivalent to the product below when [tex]x \ \textgreater \ 0[/tex]?

[tex]\sqrt{\frac{6}{x}} \cdot \sqrt{\frac{x^2}{24}}[/tex]

A. [tex]\sqrt{\frac{x}{2}}[/tex]
B. [tex]\frac{x}{4}[/tex]
C. [tex]\frac{x}{2}[/tex]
D. [tex]\frac{\sqrt{x}}{2}[/tex]


Sagot :

Let's simplify the given expression step-by-step:

The expression we are given is:
[tex]\[ \sqrt{\frac{6}{x}} \cdot \sqrt{\frac{x^2}{24}} \][/tex]

First, recall that the product of square roots is the square root of the product:
[tex]\[ \sqrt{\frac{6}{x}} \cdot \sqrt{\frac{x^2}{24}} = \sqrt{\left(\frac{6}{x}\right) \cdot \left(\frac{x^2}{24}\right)} \][/tex]

Next, we simplify the expression inside the square root. Multiply the fractions together:
[tex]\[ \left(\frac{6}{x}\right) \cdot \left(\frac{x^2}{24}\right) = \frac{6 \cdot x^2}{x \cdot 24} = \frac{6x^2}{24x} \][/tex]

Notice that the expression [tex]\(\frac{6x^2}{24x}\)[/tex] can be simplified further. Divide both the numerator and the denominator by 6:
[tex]\[ \frac{6x^2}{24x} = \frac{x^2}{4x} \][/tex]

Then simplify the fraction by canceling out [tex]\(x\)[/tex] in the numerator and the denominator:
[tex]\[ \frac{x^2}{4x} = \frac{x}{4} \][/tex]

We now have:
[tex]\[ \sqrt{\frac{6}{x}} \cdot \sqrt{\frac{x^2}{24}} = \sqrt{\frac{x}{4}} \][/tex]

Recognize that taking the square root of a fraction [tex]\(\frac{a}{b}\)[/tex] is the same as [tex]\(\frac{\sqrt{a}}{\sqrt{b}}\)[/tex]:
[tex]\[ \sqrt{\frac{x}{4}} = \frac{\sqrt{x}}{\sqrt{4}} \][/tex]

Since [tex]\(\sqrt{4} = 2\)[/tex], we can write:
[tex]\[ \frac{\sqrt{x}}{\sqrt{4}} = \frac{\sqrt{x}}{2} \][/tex]

Thus, the simplified expression is:
[tex]\[ \frac{\sqrt{x}}{2} \][/tex]

Therefore, the choice equivalent to the given product when [tex]\(x > 0\)[/tex] is:

[tex]\[ \boxed{\frac{\sqrt{x}}{2}} \][/tex]