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3. Rationalize the denominator in each expression. Express each answer in simplest form. \( \begin{array}{ll}\text { a) } \frac{2 \sqrt{3}+4}{\sqrt{3}}(\sqrt{3}) & \frac{2(3)+4 \sqrt{3}}{3}\end{array} \)

Ask by Todd Mccoy. in Canada
Dec 03,2024

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Answer

The rationalized form is \( \frac{6 + 4 \sqrt{3}}{3} \).

Solution

To rationalize the denominator in the expression \( \frac{2 \sqrt{3}+4}{\sqrt{3}} \), we need to multiply both the numerator and the denominator by \( \sqrt{3} \) to eliminate the square root in the denominator. Given expression: \( \frac{2 \sqrt{3}+4}{\sqrt{3}} \) Multiplying both the numerator and the denominator by \( \sqrt{3} \): \[ \frac{(2 \sqrt{3}+4) \cdot \sqrt{3}}{\sqrt{3} \cdot \sqrt{3}} \] Simplifying: \[ \frac{2 \cdot 3 + 4 \cdot \sqrt{3}}{3} \] \[ \frac{6 + 4 \sqrt{3}}{3} \] Therefore, the rationalized form of the expression is \( \frac{6 + 4 \sqrt{3}}{3} \).

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To rationalize the denominator in the expression \( \frac{2 \sqrt{3}+4}{\sqrt{3}} \), we multiply the numerator and the denominator by \(\sqrt{3}\): \[ \frac{(2 \sqrt{3}+4) \cdot \sqrt{3}}{\sqrt{3} \cdot \sqrt{3}} = \frac{2(3) + 4\sqrt{3}}{3} = \frac{6 + 4\sqrt{3}}{3}. \] Now, simplifying this gives us: \[ \frac{6}{3} + \frac{4\sqrt{3}}{3} = 2 + \frac{4\sqrt{3}}{3}. \] Therefore, the final simplified form is: \[ 2 + \frac{4\sqrt{3}}{3}. \]

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