finals review #1 - radicals, exponents, sequences, and key features
simplify the following radical expressions completely.
1. $\sqrt{50x^7}$
2. $3\sqrt3{48x^3y^7z}$
3. $4\sqrt{8} + 3\sqrt{50} - 6\sqrt{32}$
4. $\frac{2\sqrt{4}}{\sqrt{3}}$
5. $\sqrt{5}(-2\sqrt{10} + 2)$
6. $\frac{-2 + 4\sqrt{3}}{\sqrt{2}}$
simplify the following exponential expressions completely.
7. $(3x^2y^4)^3$
8. $4xy^{-3} \cdot 2x^{-3}y^4$
9. $\frac{-uv^3 \cdot u^4}{2v^2}$

Question

Accepted Answer

### Problem 1: Simplify $\boldsymbol{\sqrt{50x^7}}$
# Explanation:
## Step1: Factor the radicand
Factor $50$ into $25 	imes 2$ and $x^7$ into $x^6 	imes x$. So, $\sqrt{50x^7} = \sqrt{25 	imes 2 	imes x^6 	imes x}$.
## Step2: Simplify the square root
Use the property $\sqrt{ab} = \sqrt{a}\sqrt{b}$ and $\sqrt{a^2} = a$ (for $a \geq 0$). We get $\sqrt{25}\sqrt{2}\sqrt{x^6}\sqrt{x} = 5 	imes \sqrt{2} 	imes x^3 	imes \sqrt{x}$.
## Step3: Combine like terms
Multiply the constant and the variable terms: $5x^3\sqrt{2x}$.
# Answer:
$5x^3\sqrt{2x}$

### Problem 2: Simplify $\boldsymbol{3\sqrt[3]{48x^3y^7z}}$
# Explanation:
## Step1: Factor the radicand
Factor $48$ into $8 	imes 6$, $x^3$ remains as is, $y^7$ into $y^6 	imes y$, and $z$ remains as is. So, $3\sqrt[3]{48x^3y^7z} = 3\sqrt[3]{8 	imes 6 	imes x^3 	imes y^6 	imes y 	imes z}$.
## Step2: Simplify the cube root
Use the property $\sqrt[3]{ab} = \sqrt[3]{a}\sqrt[3]{b}$ and $\sqrt[3]{a^3} = a$. We get $3 	imes \sqrt[3]{8}\sqrt[3]{6}\sqrt[3]{x^3}\sqrt[3]{y^6}\sqrt[3]{y}\sqrt[3]{z} = 3 	imes 2 	imes \sqrt[3]{6} 	imes x 	imes y^2 	imes \sqrt[3]{y} 	imes \sqrt[3]{z}$.
## Step3: Combine like terms
Multiply the constant and the variable terms: $6xy^2\sqrt[3]{6yz}$.
# Answer:
$6xy^2\sqrt[3]{6yz}$

### Problem 3: Simplify $\boldsymbol{4\sqrt{8} + 3\sqrt{50} - 6\sqrt{32}}$
# Explanation:
## Step1: Simplify each radical
- For $\sqrt{8}$, factor into $\sqrt{4 	imes 2} = 2\sqrt{2}$, so $4\sqrt{8} = 4 	imes 2\sqrt{2} = 8\sqrt{2}$.
- For $\sqrt{50}$, factor into $\sqrt{25 	imes 2} = 5\sqrt{2}$, so $3\sqrt{50} = 3 	imes 5\sqrt{2} = 15\sqrt{2}$.
- For $\sqrt{32}$, factor into $\sqrt{16 	imes 2} = 4\sqrt{2}$, so $6\sqrt{32} = 6 	imes 4\sqrt{2} = 24\sqrt{2}$.
## Step2: Combine like terms
Substitute the simplified radicals back: $8\sqrt{2} + 15\sqrt{2} - 24\sqrt{2} = (8 + 15 - 24)\sqrt{2} = (23 - 24)\sqrt{2} = -\sqrt{2}$.
# Answer:
$-\sqrt{2}$

### Problem 4: Simplify $\boldsymbol{\frac{2\sqrt{4}}{\sqrt{3}}}$
# Explanation:
## Step1: Simplify $\sqrt{4}$
We know that $\sqrt{4} = 2$, so the numerator becomes $2 	imes 2 = 4$. Now the expression is $\frac{4}{\sqrt{3}}$.
## Step2: Rationalize the denominator
Multiply numerator and denominator by $\sqrt{3}$: $\frac{4 	imes \sqrt{3}}{\sqrt{3} 	imes \sqrt{3}} = \frac{4\sqrt{3}}{3}$.
# Answer:
$\frac{4\sqrt{3}}{3}$

### Problem 5: Simplify $\boldsymbol{\sqrt{5}(-2\sqrt{10} + 2)}$
# Explanation:
## Step1: Distribute $\sqrt{5}$
Use the distributive property $a(b + c) = ab + ac$. So, $\sqrt{5} 	imes (-2\sqrt{10}) + \sqrt{5} 	imes 2$.
## Step2: Simplify each term
- For $\sqrt{5} 	imes (-2\sqrt{10})$, use $\sqrt{a}\sqrt{b} = \sqrt{ab}$: $-2\sqrt{50} = -2\sqrt{25 	imes 2} = -2 	imes 5\sqrt{2} = -10\sqrt{2}$.
- For $\sqrt{5} 	imes 2$, it is $2\sqrt{5}$.
## Step3: Combine the terms
The simplified expression is $-10\sqrt{2} + 2\sqrt{5}$.
# Answer:
$-10\sqrt{2} + 2\sqrt{5}$

### Problem 6: Simplify $\boldsymbol{\frac{-2 + 4\sqrt{3}}{\sqrt{2}}}$
# Explanation:
## Step1: Rationalize the denominator
Multiply numerator and denominator by $\sqrt{2}$: $\frac{(-2 + 4\sqrt{3}) 	imes \sqrt{2}}{\sqrt{2} 	imes \sqrt{2}} = \frac{-2\sqrt{2} + 4\sqrt{6}}{2}$.
## Step2: Simplify the fraction
Divide each term in the numerator by 2: $\frac{-2\sqrt{2}}{2} + \frac{4\sqrt{6}}{2} = -\sqrt{2} + 2\sqrt{6}$.
# Answer:
$2\sqrt{6} - \sqrt{2}$ (or $-\sqrt{2} + 2\sqrt{6}$)

### Problem 7: Simplify $\boldsymbol{(3x^2y^4)^3}$
# Explanation:
## Step1: Apply the power of a product rule
Use $(ab)^n = a^n b^n$. So, $(3x^2y^4)^3 = 3^3 	imes (x^2)^3 	imes (y^4)^3$.
## Step2: Simplify each term
- $3^3 = 27$.
- $(x^2)^3 = x^{2 	imes 3} = x^6$ (using $(a^m)^n = a^{mn}$).
- $(y^4)^3 = y^{4 	imes 3} = y^{12}$.
## Step3: Combine the terms
Multiply the results: $27x^6y^{12}$.
# Answer:
$27x^6y^{12}$

### Problem 8: Simplify $\boldsymbol{4xy^{-3} \cdot 2x^{-3}y^4}$
# Explanation:
## Step1: Multiply the coefficients and use exponent rules for variables
Multiply the coefficients $4 	imes 2 = 8$. For the $x$ terms, use $a^m \cdot a^n = a^{m + n}$: $x^1 \cdot x^{-3} = x^{1 + (-3)} = x^{-2}$. For the $y$ terms: $y^{-3} \cdot y^4 = y^{-3 + 4} = y^1 = y$.
## Step2: Rewrite with positive exponents (if needed)
We can write $x^{-2}$ as $\frac{1}{x^2}$, but we can also leave it as is or combine. The expression is $8x^{-2}y$. If we want positive exponents, it is $\frac{8y}{x^2}$.
# Answer:
$\frac{8y}{x^2}$ (or $8x^{-2}y$)

### Problem 9: Simplify $\boldsymbol{\frac{-uv^3 \cdot u^4}{2v^2}}$
# Explanation:
## Step1: Simplify the numerator for $u$ and $v$ terms
For the $u$ terms: $u^1 \cdot u^4 = u^{1 + 4} = u^5$. For the $v$ terms: $v^3$ (since we have $v^3$ in the numerator and $v^2$ in the denominator, we'll handle that next).
## Step2: Simplify the fraction for $v$ terms
Use $\frac{a^m}{a^n} = a^{m - n}$: $\frac{v^3}{v^2} = v^{3 - 2} = v^1 = v$.
## Step3: Combine the terms
The numerator is $-u^5v$ and the denominator is $2$, so the expression is $\frac{-u^5v}{2}$.
# Answer:
$-\frac{u^5v}{2}$

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QUESTION IMAGE

Question

Explanation:

Problem 1: Simplify $\boldsymbol{\sqrt{50x^7}}$

Step1: Factor the radicand

Step2: Simplify the square root

Step3: Combine like terms

Step1: Factor the radicand

Step2: Simplify the cube root

Step3: Combine like terms

Step1: Simplify each radical

Step2: Combine like terms

Answer:

Problem 2: Simplify $\boldsymbol{3\sqrt[3]{48x^3y^7z}}$