Rationalize the denominator. Assume that all variables represent positive real numbers and that the denominator is not zero.
-$\frac{4}{6-\sqrt{3}}$

A) $\frac{24+4 \sqrt{3}}{33}$

B) $\frac{24+4 \sqrt{3}}{3}$

C) $\frac{4}{6}-\frac{4}{\sqrt{3}}$

D) $\frac{24-4 \sqrt{3}}{33}$

Question

Rationalize the denominator. Assume that all variables represent positive real numbers and that the denominator is not zero.
-$\frac{4}{6-\sqrt{3}}$

A) $\frac{24+4 \sqrt{3}}{33}$

B) $\frac{24+4 \sqrt{3}}{3}$

C) $\frac{4}{6}-\frac{4}{\sqrt{3}}$

D) $\frac{24-4 \sqrt{3}}{33}$

Quizplus · Accepted Answer

To rationalize the denominator of $\frac{4}{6-\sqrt{3}}$, multiply both the numerator and the denominator by the conjugate of the denominator, which is $6+\sqrt{3}$. This gives $\frac{4(6+\sqrt{3})}{(6-\sqrt{3})(6+\sqrt{3})} = \frac{24+4\sqrt{3}}{36-3} = \frac{24+4\sqrt{3}}{33}$.

Rationalize the Denominator 46−3\frac{4}{6-\sqrt{3}}6−3​4​ A) 24+4333\frac{24+4 \sqrt{3}}{33}3324+43​​ B) 24+433\frac{24+4 \sqrt{3}}{3}324+43​​ C) 46−43\frac{4}{6}-\frac{4}{\sqrt{3}}64​−3​4​ D) 24−4333\frac{24-4 \sqrt{3}}{33}3324−43​​

Rationalize the Denominator $\frac{4}{6-\sqrt{3}}$ A) $\frac{24+4 \sqrt{3}}{33}$ B) $\frac{24+4 \sqrt{3}}{3}$ C) $\frac{4}{6}-\frac{4}{\sqrt{3}}$ D) $\frac{24-4 \sqrt{3}}{33}$