Algebra 1B

Before attempting to solve this expression, it is important to simplify it. We can use the property that $\sqrt{a}*\sqrt{b} = \sqrt{ab}$ to simplify the expression. \begin{align*} \sqrt{\frac{33xy^{3}}{\sqrt{3x}}} &= \sqrt{\frac{33xy^{3}*3x}{3x}}\\ &=\sqrt{33xy^3*3x}\\ &=\sqrt{99x^2y^3} \end{align*} Now that the expression is simplified, we can solve it by multiplying together the factors under the square root. \begin{align*} \sqrt{99x^2y^3} &= \sqrt{11x*9xy^2}\\ &=11x\sqrt{9xy^2} \end{align*} Since the last part of the expression is still under a square root, we can use the fact that $\sqrt{ab} = \sqrt{a}\sqrt{b}$ again to get: \begin{align*} 11x\sqrt{9xy^2} &= 11x\sqrt{9x}\sqrt{y^2}