Question

8.5 divide radical expressions (homework)
score: 9/14 answered: 10/14
question 11
simplify.
\\(\frac{\sqrt{64c^{20}y^{5}}}{\sqrt{2c^{5}y^{15}}}=\\)
question help: video written example

Explanation:

Step1: Use quotient - rule of radicals

The quotient - rule states that $\frac{\sqrt{a}}{\sqrt{b}}=\sqrt{\frac{a}{b}}$ for $a\geq0$ and $b > 0$. So, $\frac{\sqrt{64c^{20}y^{5}}}{\sqrt{2c^{5}y^{15}}}=\sqrt{\frac{64c^{20}y^{5}}{2c^{5}y^{15}}}$.

Step2: Simplify the fraction inside the radical

First, simplify the coefficient: $\frac{64}{2} = 32$. Then, use the rule of exponents $\frac{x^{m}}{x^{n}}=x^{m - n}$ for $x
eq0$. For the $c$ terms, $\frac{c^{20}}{c^{5}}=c^{20 - 5}=c^{15}$, and for the $y$ terms, $\frac{y^{5}}{y^{15}}=y^{5-15}=y^{- 10}=\frac{1}{y^{10}}$. So, $\sqrt{\frac{64c^{20}y^{5}}{2c^{5}y^{15}}}=\sqrt{32c^{15}y^{-10}}$.

Step3: Rewrite the radical as a product of radicals

$\sqrt{32c^{15}y^{-10}}=\sqrt{32}\cdot\sqrt{c^{15}}\cdot\sqrt{y^{-10}}$.

Step4: Simplify each radical

$\sqrt{32}=\sqrt{16\times2}=4\sqrt{2}$, $\sqrt{c^{15}}=c^{7}\sqrt{c}$ (since $15 = 2\times7+1$), and $\sqrt{y^{-10}}=\frac{1}{y^{5}}$.

Step5: Combine the simplified radicals

$4\sqrt{2}\cdot c^{7}\sqrt{c}\cdot\frac{1}{y^{5}}=\frac{4c^{7}\sqrt{2c}}{y^{5}}$.

Answer:

$\frac{4c^{7}\sqrt{2c}}{y^{5}}$