elementary algebra skill
simplifying radicals of index 2 (assume all variables are positive.)
simplify.
1) $\sqrt{125}$
2) $\sqrt{72}$
3) $\sqrt{63}$
4) $\sqrt{45}$
5) $\sqrt{12}$
6) $\sqrt{18}$
7) $\sqrt{96}$
8) $\sqrt{150}$
9) $\sqrt{112}$
10) $\sqrt{50}$
11) $\sqrt{8a^4}$
12) $\sqrt{200r}$
13) $\sqrt{20x^3}$
14) $\sqrt{45n}$
15) $\sqrt{98x^5}$
16) $\sqrt{24a^4}$
17) $\sqrt{75x^5}$
18) $\sqrt{32n^7}$
19) $\sqrt{98n^{12}}$
20) $\sqrt{28k^9}$
21) $\sqrt{18x^4y}$
22) $\sqrt{24mn^2}$
23) $\sqrt{50x^3y^4}$
24) $\sqrt{320mn^4}$
25) $\sqrt{192a^3b}$
26) $\sqrt{80x^2y^4}$
27) $\sqrt{112x^3y^4}$
28) $\sqrt{256x^4y^4}$
29) $\sqrt{216x^3y^3}$
30) $\sqrt{192u^4v^4}$

Question

Accepted Answer

Let's solve problem 1: $\boldsymbol{\sqrt{125}}$

# Explanation:
## Step1: Factor 125 into perfect square and other
We know that $125 = 25	imes5$, where $25$ is a perfect square ($5^2$). So we can rewrite the square root as:
$\sqrt{125}=\sqrt{25	imes5}$

## Step2: Use the property of square roots $\sqrt{ab}=\sqrt{a}	imes\sqrt{b}$ (for $a\geq0,b\geq0$)
Applying the property, we get:
$\sqrt{25	imes5}=\sqrt{25}	imes\sqrt{5}$

## Step3: Simplify $\sqrt{25}$
Since $\sqrt{25} = 5$ (because $5^2 = 25$), we substitute that in:
$5	imes\sqrt{5}=5\sqrt{5}$

# Answer:
$5\sqrt{5}$

Let's solve problem 2: $\boldsymbol{\sqrt{72}}$

# Explanation:
## Step1: Factor 72 into perfect square and other
We know that $72 = 36	imes2$, where $36$ is a perfect square ($6^2$). So we can rewrite the square root as:
$\sqrt{72}=\sqrt{36	imes2}$

## Step2: Use the property of square roots $\sqrt{ab}=\sqrt{a}	imes\sqrt{b}$ (for $a\geq0,b\geq0$)
Applying the property, we get:
$\sqrt{36	imes2}=\sqrt{36}	imes\sqrt{2}$

## Step3: Simplify $\sqrt{36}$
Since $\sqrt{36} = 6$ (because $6^2 = 36$), we substitute that in:
$6	imes\sqrt{2}=6\sqrt{2}$

# Answer:
$6\sqrt{2}$

Let's solve problem 3: $\boldsymbol{\sqrt{63}}$

# Explanation:
## Step1: Factor 63 into perfect square and other
We know that $63 = 9	imes7$, where $9$ is a perfect square ($3^2$). So we can rewrite the square root as:
$\sqrt{63}=\sqrt{9	imes7}$

## Step2: Use the property of square roots $\sqrt{ab}=\sqrt{a}	imes\sqrt{b}$ (for $a\geq0,b\geq0$)
Applying the property, we get:
$\sqrt{9	imes7}=\sqrt{9}	imes\sqrt{7}$

## Step3: Simplify $\sqrt{9}$
Since $\sqrt{9} = 3$ (because $3^2 = 9$), we substitute that in:
$3	imes\sqrt{7}=3\sqrt{7}$

# Answer:
$3\sqrt{7}$

Let's solve problem 4: $\boldsymbol{\sqrt{45}}$

# Explanation:
## Step1: Factor 45 into perfect square and other
We know that $45 = 9	imes5$, where $9$ is a perfect square ($3^2$). So we can rewrite the square root as:
$\sqrt{45}=\sqrt{9	imes5}$

## Step2: Use the property of square roots $\sqrt{ab}=\sqrt{a}	imes\sqrt{b}$ (for $a\geq0,b\geq0$)
Applying the property, we get:
$\sqrt{9	imes5}=\sqrt{9}	imes\sqrt{5}$

## Step3: Simplify $\sqrt{9}$
Since $\sqrt{9} = 3$ (because $3^2 = 9$), we substitute that in:
$3	imes\sqrt{5}=3\sqrt{5}$

# Answer:
$3\sqrt{5}$

Let's solve problem 5: $\boldsymbol{\sqrt{12}}$

# Explanation:
## Step1: Factor 12 into perfect square and other
We know that $12 = 4	imes3$, where $4$ is a perfect square ($2^2$). So we can rewrite the square root as:
$\sqrt{12}=\sqrt{4	imes3}$

## Step2: Use the property of square roots $\sqrt{ab}=\sqrt{a}	imes\sqrt{b}$ (for $a\geq0,b\geq0$)
Applying the property, we get:
$\sqrt{4	imes3}=\sqrt{4}	imes\sqrt{3}$

## Step3: Simplify $\sqrt{4}$
Since $\sqrt{4} = 2$ (because $2^2 = 4$), we substitute that in:
$2	imes\sqrt{3}=2\sqrt{3}$

# Answer:
$2\sqrt{3}$

Let's solve problem 6: $\boldsymbol{\sqrt{18}}$

# Explanation:
## Step1: Factor 18 into perfect square and other
We know that $18 = 9	imes2$, where $9$ is a perfect square ($3^2$). So we can rewrite the square root as:
$\sqrt{18}=\sqrt{9	imes2}$

## Step2: Use the property of square roots $\sqrt{ab}=\sqrt{a}	imes\sqrt{b}$ (for $a\geq0,b\geq0$)
Applying the property, we get:
$\sqrt{9	imes2}=\sqrt{9}	imes\sqrt{2}$

## Step3: Simplify $\sqrt{9}$
Since $\sqrt{9} = 3$ (because $3^2 = 9$), we substitute that in:
$3	imes\sqrt{2}=3\sqrt{2}$

# Answer:
$3\sqrt{2}$

Let's solve problem 7: $\boldsymbol{\sqrt{96}}$

# Explanation:
## Step1: Factor 96 into perfect square and other
We know that $96 = 16	imes6$, where $16$ is a perfect square ($4^2$). So we can rewrite the square root as:
$\sqrt{96}=\sqrt{16	imes6}$

## Step2: Use the property of square roots $\sqrt{ab}=\sqrt{a}	imes\sqrt{b}$ (for $a\geq0,b\geq0$)
Applying the property, we get:
$\sqrt{16	imes6}=\sqrt{16}	imes\sqrt{6}$

## Step3: Simplify $\sqrt{16}$
Since $\sqrt{16} = 4$ (because $4^2 = 16$), we substitute that in:
$4	imes\sqrt{6}=4\sqrt{6}$

# Answer:
$4\sqrt{6}$

Let's solve problem 8: $\boldsymbol{\sqrt{150}}$

# Explanation:
## Step1: Factor 150 into perfect square and other
We know that $150 = 25	imes6$, where $25$ is a perfect square ($5^2$). So we can rewrite the square root as:
$\sqrt{150}=\sqrt{25	imes6}$

## Step2: Use the property of square roots $\sqrt{ab}=\sqrt{a}	imes\sqrt{b}$ (for $a\geq0,b\geq0$)
Applying the property, we get:
$\sqrt{25	imes6}=\sqrt{25}	imes\sqrt{6}$

## Step3: Simplify $\sqrt{25}$
Since $\sqrt{25} = 5$ (because $5^2 = 25$), we substitute that in:
$5	imes\sqrt{6}=5\sqrt{6}$

# Answer:
$5\sqrt{6}$

Let's solve problem 9: $\boldsymbol{\sqrt{112}}$

# Explanation:
## Step1: Factor 112 into perfect square and other
We know that $112 = 16	imes7$, where $16$ is a perfect square ($4^2$). So we can rewrite the square root as:
$\sqrt{112}=\sqrt{16	imes7}$

## Step2: Use the property of square roots $\sqrt{ab}=\sqrt{a}	imes\sqrt{b}$ (for $a\geq0,b\geq0$)
Applying the property, we get:
$\sqrt{16	imes7}=\sqrt{16}	imes\sqrt{7}$

## Step3: Simplify $\sqrt{16}$
Since $\sqrt{16} = 4$ (because $4^2 = 16$), we substitute that in:
$4	imes\sqrt{7}=4\sqrt{7}$

# Answer:
$4\sqrt{7}$

Let's solve problem 10: $\boldsymbol{\sqrt{50}}$

# Explanation:
## Step1: Factor 50 into perfect square and other
We know that $50 = 25	imes2$, where $25$ is a perfect square ($5^2$). So we can rewrite the square root as:
$\sqrt{50}=\sqrt{25	imes2}$

## Step2: Use the property of square roots $\sqrt{ab}=\sqrt{a}	imes\sqrt{b}$ (for $a\geq0,b\geq0$)
Applying the property, we get:
$\sqrt{25	imes2}=\sqrt{25}	imes\sqrt{2}$

## Step3: Simplify $\sqrt{25}$
Since $\sqrt{25} = 5$ (because $5^2 = 25$), we substitute that in:
$5	imes\sqrt{2}=5\sqrt{2}$

# Answer:
$5\sqrt{2}$

Let's solve problem 11: $\boldsymbol{\sqrt{8a^4}}$

# Explanation:
## Step1: Factor the radicand
We can factor $8a^4$ as $4a^4	imes2$, where $4a^4$ is a perfect square (since $(2a^2)^2 = 4a^4$). So we rewrite the square root:
$\sqrt{8a^4}=\sqrt{4a^4	imes2}$

## Step2: Use the property of square roots $\sqrt{ab}=\sqrt{a}	imes\sqrt{b}$ (for $a\geq0,b\geq0$)
Applying the property, we get:
$\sqrt{4a^4	imes2}=\sqrt{4a^4}	imes\sqrt{2}$

## Step3: Simplify $\sqrt{4a^4}$
Since $\sqrt{4a^4} = 2a^2$ (because $(2a^2)^2 = 4a^4$), we substitute that in:
$2a^2	imes\sqrt{2}=2a^2\sqrt{2}$

# Answer:
$2a^2\sqrt{2}$

Let's solve problem 12: $\boldsymbol{\sqrt{200r}}$

# Explanation:
## Step1: Factor the radicand
We can factor $200r$ as $100	imes2r$, where $100$ is a perfect square ($10^2$). So we rewrite the square root:
$\sqrt{200r}=\sqrt{100	imes2r}$

## Step2: Use the property of square roots $\sqrt{ab}=\sqrt{a}	imes\sqrt{b}$ (for $a\geq0,b\geq0$)
Applying the property, we get:
$\sqrt{100	imes2r}=\sqrt{100}	imes\sqrt{2r}$

## Step3: Simplify $\sqrt{100}$
Since $\sqrt{100} = 10$ (because $10^2 = 100$), we substitute that in:
$10	imes\sqrt{2r}=10\sqrt{2r}$

# Answer:
$10\sqrt{2r}$

Let's solve problem 13: $\boldsymbol{\sqrt{20x^3}}$

# Explanation:
## Step1: Factor the radicand
We can factor $20x^3$ as $4x^2	imes5x$, where $4x^2$ is a perfect square (since $(2x)^2 = 4x^2$). So we rewrite the square root:
$\sqrt{20x^3}=\sqrt{4x^2	imes5x}$

## Step2: Use the property of square roots $\sqrt{ab}=\sqrt{a}	imes\sqrt{b}$ (for $a\geq0,b\geq0$)
Applying the property, we get:
$\sqrt{4x^2	imes5x}=\sqrt{4x^2}	imes\sqrt{5x}$

## Step3: Simplify $\sqrt{4x^2}$
Since $\sqrt{4x^2} = 2x$ (because $(2x)^2 = 4x^2$ and $x\gt0$), we substitute that in:
$2x	imes\sqrt{5x}=2x\sqrt{5x}$

# Answer:
$2x\sqrt{5x}$

Let's solve problem 14: $\boldsymbol{\sqrt{45n}}$

# Explanation:
## Step1: Factor the radicand
We can factor $45n$ as $9	imes5n$, where $9$ is a perfect square ($3^2$). So we rewrite the square root:
$\sqrt{45n}=\sqrt{9	imes5n}$

## Step2: Use the property of square roots $\sqrt{ab}=\sqrt{a}	imes\sqrt{b}$ (for $a\geq0,b\geq0$)
Applying the property, we get:
$\sqrt{9	imes5n}=\sqrt{9}	imes\sqrt{5n}$

## Step3: Simplify $\sqrt{9}$
Since $\sqrt{9} = 3$ (because $3^2 = 9$), we substitute that in:
$3	imes\sqrt{5n}=3\sqrt{5n}$

# Answer:
$3\sqrt{5n}$

Let's solve problem 15: $\boldsymbol{\sqrt{98x^5}}$

# Explanation:
## Step1: Factor the radicand
We can factor $98x^5$ as $49x^4	imes2x$, where $49x^4$ is a perfect square (since $(7x^2)^2 = 49x^4$). So we rewrite the square root:
$\sqrt{98x^5}=\sqrt{49x^4	imes2x}$

## Step2: Use the property of square roots $\sqrt{ab}=\sqrt{a}	imes\sqrt{b}$ (for $a\geq0,b\geq0$)
Applying the property, we get:
$\sqrt{49x^4	imes2x}=\sqrt{49x^4}	imes\sqrt{2x}$

## Step3: Simplify $\sqrt{49x^4}$
Since $\sqrt{49x^4} = 7x^2$ (because $(7x^2)^2 = 49x^4$ and $x\gt0$), we substitute that in:
$7x^2	imes\sqrt{2x}=7x^2\sqrt{2x}$

# Answer:
$7x^2\sqrt{2x}$

Let's solve problem 16: $\boldsymbol{\sqrt{24a^4}}$

# Explanation:
## Step1: Factor the radicand
We can factor $24a^4$ as $4a^4	imes6$, where $4a^4$ is a perfect square (since $(2a^2)^2 = 4a^4$). So we rewrite the square root:
$\sqrt{24a^4}=\sqrt{4a^4	imes6}$

## Step2: Use the property of square roots $\sqrt{ab}=\sqrt{a}	imes\sqrt{b}$ (for $a\geq0,b\geq0$)
Applying the property, we get:
$\sqrt{4a^4	imes6}=\sqrt{4a^4}	imes\sqrt{6}$

## Step3: Simplify $\sqrt{4a^4}$
Since $\sqrt{4a^4} = 2a^2$ (because $(2a^2)^2 = 4a^4$ and $a\gt0$), we substitute that in:
$2a^2	imes\sqrt{6}=2a^2\sqrt{6}$

# Answer:
$2a^2\sqrt{6}$

Let's solve problem 17: $\boldsymbol{\sqrt{75x^5}}$

# Explanation:
## Step1: Factor the radicand
We can factor $75x^5$ as $25x^4	imes3x$, where $25x^4$ is a perfect square (since $(5x^2)^2 = 25x^4$). So we rewrite the square root:
$\sqrt{75x^5}=\sqrt{25x^4	imes3x}$

## Step2: Use the property of square roots $\sqrt{ab}=\sqrt{a}	imes\sqrt{b}$ (for $a\geq0,b\geq0$)
Applying the property, we get:
$\sqrt{25x^4	imes3x}=\sqrt{25x^4}	imes\sqrt{3x}$

## Step3: Simplify $\sqrt{25x^4}$
Since $\sqrt{25x^4} = 5x^2$ (because $(5x^2)^2 = 25x^4$ and $x\gt0$), we substitute that in:
$5x^2	imes\sqrt{3x}=5x^2\sqrt{3x}$

# Answer:
$5x^2\sqrt{3x}$

Let's solve problem 18: $\boldsymbol{\sqrt{32n^7}}$

# Explanation:
## Step1: Factor the radic

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QUESTION IMAGE

Question

Explanation:

Step1: Factor 125 into perfect square and other

Step2: Use the property of square roots $\sqrt{ab}=\sqrt{a}\times\sqrt{b}$ (for $a\geq0,b\geq0$)

Step3: Simplify $\sqrt{25}$

Step1: Factor 72 into perfect square and other

Step2: Use the property of square roots $\sqrt{ab}=\sqrt{a}\times\sqrt{b}$ (for $a\geq0,b\geq0$)

Step3: Simplify $\sqrt{36}$

Step1: Factor 63 into perfect square and other

Step2: Use the property of square roots $\sqrt{ab}=\sqrt{a}\times\sqrt{b}$ (for $a\geq0,b\geq0$)

Step3: Simplify $\sqrt{9}$

Answer: