name:
pre-calculus 11: chapter 5 quiz – calculator
date:
1) simplify radicals and collect like terms. state any restrictions on the values for the variables.
a. $4\sqrt{45x^3} - \sqrt{27x} + 17\sqrt{3x} - 9\sqrt{125x^3}$
$x\geq0$
2) expand and simplify. state any restrictions on the values for the variables.
a. $(10r - 4\sqrt3{4r})(2\sqrt3{6r^2} + 3\sqrt3{12r})$
$r\geq0$
3) simplify.
a. $\frac{6\sqrt3{4v^7}}{\sqrt3{14v}}, v 0$

Question

name:
pre-calculus 11: chapter 5 quiz – calculator
date:
1) simplify radicals and collect like terms. state any restrictions on the values for the variables.
a. $4\sqrt{45x^3} - \sqrt{27x} + 17\sqrt{3x} - 9\sqrt{125x^3}$
$x\geq0$
2) expand and simplify. state any restrictions on the values for the variables.
a. $(10r - 4\sqrt3{4r})(2\sqrt3{6r^2} + 3\sqrt3{12r})$
$r\geq0$
3) simplify.
a. $\frac{6\sqrt3{4v^7}}{\sqrt3{14v}}, v > 0$

Accepted Answer

r^2} \)).

## Step 4: State restrictions
For cube roots, $ r $ can be any real number (cube roots are defined for all real numbers).

# Answer:
Expanded and simplified: $ 20r\sqrt[3]{6r^2} + 30r\sqrt[3]{12r} - 16r\sqrt[3]{3} - 24\sqrt[3]{6r^2} $; Restriction: $ r \in \mathbb{R} $ (all real numbers)

### 3a) Simplify $ \frac{\sqrt[6]{49v^7}}{\sqrt[3]{14v}} $ with $ v > 0 $.

# Explanation:
## Step 1: Rewrite radicals as exponents (optional, but helpful)
$ \sqrt[6]{49v^7} = (49v^7)^{\frac{1}{6}} = 49^{\frac{1}{6}}v^{\frac{7}{6}} $.
$ \sqrt[3]{14v} = (14v)^{\frac{1}{3}} = 14^{\frac{1}{3}}v^{\frac{1}{3}} $.

## Step 2: Rewrite with common denominator for exponents
$ v^{\frac{7}{6}} \div v^{\frac{1}{3}} = v^{\frac{7}{6} - \frac{2}{6}} = v^{\frac{5}{6}} $.

Rewrite $ 49^{\frac{1}{6}} $ as $ (7^2)^{\frac{1}{6}} = 7^{\frac{1}{3}} $.
Rewrite $ 14^{\frac{1}{3}} = (2 \cdot 7)^{\frac{1}{3}} = 2^{\frac{1}{3}} \cdot 7^{\frac{1}{3}} $.

## Step 3: Simplify the coefficient
$ \frac{7^{\frac{1}{3}}}{2^{\frac{1}{3}} \cdot 7^{\frac{1}{3}}} = \frac{1}{2^{\frac{1}{3}}} $ (the $ 7^{\frac{1}{3}} $ terms cancel).

## Step 4: Combine back
Now we have $ \frac{1}{2^{\frac{1}{3}}} \cdot v^{\frac{5}{6}} $. To rationalize or simplify, rewrite $ \frac{1}{2^{\frac{1}{3}}} = 2^{-\frac{1}{3}} $, but alternatively, rewrite the original expression using radicals:

Original expression: $ \frac{\sqrt[6]{49v^7}}{\sqrt[3]{14v}} = \frac{\sqrt[6]{49v^7}}{\sqrt[6]{(14v)^2}} $ (since $ \sqrt[3]{x} = \sqrt[6]{x^2} $).

Simplify inside the sixth root: $ \frac{49v^7}{(14v)^2} = \frac{49v^7}{196v^2} = \frac{v^5}{4} $.

Thus, $ \sqrt[6]{\frac{v^5}{4}} = \frac{\sqrt[6]{v^5}}{\sqrt[6]{4}} = \frac{v^{\frac{5}{6}}}{4^{\frac{1}{6}}} $.

Rationalize $ 4^{\frac{1}{6}} = (2^2)^{\frac{1}{6}} = 2^{\frac{1}{3}} $, so multiply numerator and denominator by $ 2^{\frac{2}{3}} $:

$ \frac{v^{\frac{5}{6}} \cdot 2^{\frac{2}{3}}}{2^{\frac{1}{3}} \cdot 2^{\frac{2}{3}}} = \frac{v^{\frac{5}{6}} \cdot 2^{\frac{2}{3}}}{2} $.

But a simpler approach: Rewrite $ \sqrt[6]{49v^7} = \sqrt[6]{49v^6 \cdot v} = v\sqrt[6]{49v} $, and $ \sqrt[3]{14v} = \sqrt[6]{(14v)^2} = \sqrt[6]{196v^2} $.

Then $ \frac{v\sqrt[6]{49v}}{\sqrt[6]{196v^2}} = v \cdot \sqrt[6]{\frac{49v}{196v^2}} = v \cdot \sqrt[6]{\frac{1}{4v}} $.

Wait, earlier steps had an error. Let’s correct:

$ \sqrt[6]{49v^7} = \sqrt[6]{49v^6 \cdot v} = v\sqrt[6]{49v} $ (since $ v > 0 $, $ \sqrt[6]{v^6} = v $).

$ \sqrt[3]{14v} = \sqrt[6]{(14v)^2} = \sqrt[6]{196v^2} $.

Thus, $ \frac{v\sqrt[6]{49v}}{\sqrt[6]{196v^2}} = v \cdot \sqrt[6]{\frac{49v}{196v^2}} = v \cdot \sqrt[6]{\frac{1}{4v}} = v \cdot \frac{1}{\sqrt[6]{4v}} $.

Rationalize the denominator: Multiply numerator and denominator by $ \sqrt[6]{(4v)^5} $:

$ v \cdot \frac{\sqrt[6]{(4v)^5}}{\sqrt[6]{(4v)^6}} = v \cdot \frac{\sqrt[6]{1024v^5}}{4v} = \frac{\sqrt[6]{1024v^5}}{4} $.

Simplify $ 1024 = 2^{10} $, so $ \sqrt[6]{2^{10}v^5} = 2^{\frac{10}{6}}v^{\frac{5}{6}} = 2^{\frac{5}{3}}v^{\frac{5}{6}} $.

Thus, $ \frac{2^{\frac{5}{3}}v^{\frac{5}{6}}}{4} = \frac{2^{\frac{5}{3}}v^{\frac{5}{6}}}{2^2} = 2^{\frac{5}{3} - 2}v^{\frac{5}{6}} = 2^{-\frac{1}{3}}v^{\frac{5}{6}} $.

Alternatively, rewrite $ 2^{-\frac{1}{3}} = \frac{1}{\sqrt[3]{2}} $, so $ \frac{v^{\frac{5}{6}}}{\sqrt[3]{2}} $. Multiply numerator and denominator by $ \sqrt[3]{4} $ to rationalize:

$ \frac{v^{\frac{5}{6}} \cdot \sqrt[3]{4}}{\sqrt[3]{2} \cdot \sqrt[3]{4}} = \frac{v^{\frac{5}{6}} \cdot \sqrt[3]{4}}{\sqrt[3]{8}} = \frac{v^{\frac{5}{6}} \cdot \sqrt[3]{4}}{2} $.

But the simplest form (after correcting earlier errors) is:

Starting over:

$ \frac{\sqrt[6]{49v^7}}{\sqrt[3]{14v}} = \frac{(49v^7)^{\frac{1}{6}}}{(14v)^{\frac{1}{3}}} = \frac{49^{\frac{1}{6}}v^{\frac{7}{6}}}{14^{\frac{1}{3}}v^{\frac{1}{3}}} = \frac{7^{\frac{2}{6}}v^{\frac{7}{6}}}{(2 \cdot 7)^{\frac{1}{3}}v^{\frac{2}{6}}} = \frac{7^{\frac{1}{3}}v^{\frac{7}{6}}}{2^{\frac{1}{3}} \cdot 7^{\frac{1}{3}}v^{\frac{2}{6}}} $.

Cancel $ 7^{\frac{1}{3}} $: $ \frac{v^{\frac{7}{6} - \frac{2}{6}}}{2^{\frac{1}{3}}} = \frac{v^{\frac{5}{6}}}{2^{\frac{1}{3}}} $.

Rewrite $ v^{\frac{5}{6}} = \sqrt[6]{v^5} $ and $ 2^{\frac{1}{3}} = \sqrt[3]{2} $, so $ \frac{\sqrt[6]{v^5}}{\sqrt[3]{2}} $. Multiply numerator and denominator by $ \sqrt[3]{4} $ (since $ \sqrt[3]{2} \cdot \sqrt[3]{4} = \sqrt[3]{8} = 2 $):

$ \frac{\sqrt[6]{v^5} \cdot \sqrt[3]{4}}{2} $.

But $ \sqrt[3]{4} = 4^{\frac{1}{3}} = (2^2)^{\frac{1}{3}} = 2^{\frac{2}{3}} $, and $ \sqrt[6]{v^5} = v^{\frac{5}{6}} $, so:

$ \frac{v^{\frac{5}{6}} \cdot 2^{\frac{2}{3}}}{2} = \frac{v^{\frac{5}{6}}}{2^{1 - \frac{2}{3}}} = \frac{v^{\frac{5}{6}}}{2^{\frac{1}{3}}} $, which matches earlier.

To write with a single radical, note that $ \frac{1}{2^{\frac{1}{3}}} = 2^{-\frac{1}{3}} = (2^2)^{-\frac{1}{6}} = 4^{-\frac{1}{6}} $, so:

$ \frac{v^{\frac{5}{6}}}{4^{\frac{1}{6}}} = \sqrt[6]{\frac{v^5}{4}} $.

But the most simplified form (after correcting the initial handwritten work) is:

$ \frac{v\sqrt[6]{49v}}{\sqrt[6]{196v^2}} = \sqrt[6]{\frac{49v^7}{196v^2}} = \sqrt[6]{\frac{v^5}{4}} = \frac{\sqrt[6]{v^5}}{\sqrt[6]{4}} = \frac{v^{\frac{5}{6}}}{4^{\frac{1}{6}}} $.

Multiply numerator and denominator by $ 4^{\frac{5}{6}} $ to rationalize:

$ \frac{v^{\frac{5}{6}} \cdot 4^{\frac{5}{6}}}{4^{\frac{1}{6}} \cdot 4^{\frac{5}{6}}} = \frac{(v \cdot 4)^{\frac{5}{6}}}{4} = \frac{(4v)^{\frac{5}{6}}}{4} $.

But this is more complex. The key step is recognizing that $ \sqrt[6]{49v^7} = v\sqrt[6]{49v} $ (since $ v > 0 $) and $ \sqrt[3]{14v} = \sqrt[6]{(14v)^2} = \sqrt[6]{196v^2} $, so:

$ \frac{v\sqrt[6]{49v}}{\sqrt[6]{196v^2}} = v \cdot \sqrt[6]{\frac{49v}{196v^2}} = v \cdot \sqrt[6]{\frac{1}{4v}} = \frac{v}{\sqrt[6]{4v}} $.

Rationalizing the denominator: Multiply numerator and denominator by $ \sqrt[6]{(4v)^5} $:

$ \frac{v \cdot \sqrt[6]{(4v)^5}}{\sqrt[6]{(4v)^6}} = \frac{v \cdot \sqrt[6]{1024v^5}}{4v} = \frac{\sqrt[6]{1024v^5}}{4} $.

Simplify $ 1024 = 2^{10} $, so $ \sqrt[6]{2^{10}v^5} = 2^{\frac{10}{6}}v^{\frac{5}{6}} = 2^{\frac{5}{3}}v^{\frac{5}{6}} $.

Thus, \( \frac{2^{\frac{5}{3}}v^{\frac{5}{6}}}{4} = \frac{2^{\frac{5}{3}}v^{\frac{5}{6}}}{2^2} = 2^{\frac{5}{3} - 2}v^{\frac{5}{6

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QUESTION IMAGE

Question

Explanation:

Step 4: State restrictions

Step 1: Rewrite radicals as exponents (optional, but helpful)

Step 2: Rewrite with common denominator for exponents

Step 3: Simplify the coefficient

Step 4: Combine back

Answer:

3a) Simplify \( \frac{\sqrt[6]{49v^7}}{\sqrt[3]{14v}} \) with \( v > 0 \).