.
Back to the class
Quiz 24
1.) Simplify $\sqrt{100x^2}$ (assuming $x \geq 0$).
Solution: Using the rule that $\sqrt{A} \sqrt{B} = \sqrt{A \cdot B}$, calculate
$$\sqrt{100x^2} = \sqrt{100} \sqrt{x^2} = 10x.$$
2.) Write as one square root:
$$\left( \sqrt{3} \right) \left( \sqrt{7} \right).$$
Solution: Using the rule $\sqrt{A} \sqrt{B} = \sqrt{A \cdot B}$, calculate
$$(\sqrt{3})(\sqrt{7}) = \sqrt{3 \cdot 7} = \sqrt{21}.$$
3.) Simplify
$$\sqrt{ \dfrac{36y^2}{4x^4}}.$$
Solution: Using the rules $\sqrt{A}\sqrt{B}=\sqrt{A \cdot B}$ and $\dfrac{\sqrt{A}}{\sqrt{B}}=\sqrt{\dfrac{A}{B}}$, calculate
$$\begin{array}{ll}
\sqrt{\dfrac{36}y^2}{4x^2} &= \dfrac{\sqrt{36y^2}}{\sqrt{4x^4}} \\
&= \dfrac{\sqrt{36}\sqrt{y^2}}{\sqrt{4}\sqrt{x^4}} \\
&= \dfrac{6y}{2x^2}.
\end{array}$$
4.) Simplify (assuming $x \geq 0$)
$$\sqrt{x^{15}}.$$
Solution: Calculate
$$\sqrt{x^{15}}=\sqrt{x \cdot x^{14}} = \sqrt{x} \sqrt{x^{14}} = \sqrt{x} \sqrt{\left( x^7 \right)^2}= x^7 \sqrt{x}.$$
5.) Simplify (assuming $x \geq 0$)
$$\sqrt{ \dfrac{48x^3}{12x^5}}.$$
Solution: Calculate
$$\begin{array}{ll}
\sqrt{ \dfrac{48x^3}{12x^5} } &= \dfrac{\sqrt{48}\sqrt{x^3}}{\sqrt{12}\sqrt{x^5}} \\
&= \dfrac{\sqrt{4^2 \cdot 3} \sqrt{x} \sqrt{x^2}}{\sqrt{2^2 \cdot 3} \sqrt{x} \sqrt{x^4}} \\
&=\dfrac{4\sqrt{3}\sqrt{x} x}{2\sqrt{3}\sqrt{x} x^2} \\
&=\dfrac{4x}{2x^2} \\
&= \dfrac{2}{x}.
\end{array}$$
