operations with radicals
simplify: $\sqrt{49x^4y^8} = 7\sqrt{x^4y^8} = 7x^2y^4$
multiply: $\sqrt{3a} \cdot \sqrt{4a} = \sqrt{12a^2} = \sqrt{3 \cdot 4 \cdot a^2} = 2a\sqrt{3}$
divide: $\sqrt{\frac{18}{2}} = \sqrt{9} = 3$ $\sqrt{\frac{9}{25}} = \frac{\sqrt{9}}{\sqrt{25}} = \frac{3}{5}$
add or subtract: $2\sqrt{y} + 3\sqrt{y} + \sqrt{y} = 6\sqrt{y}$ $\sqrt{4x} + 3\sqrt{x} = 2\sqrt{x} + 3\sqrt{x} = 3\sqrt{x}$
simplify.
1. $\sqrt{x^2y^{10}} =$
2. $\sqrt{27x^8} =$
3. $\sqrt{x^{16}} =$
4. $\sqrt{125b^{15}} =$
5. $\sqrt{x^2y^4} \cdot 2\sqrt{xy} =$
6. $\sqrt{9} \cdot \sqrt{32} =$
7. $3\sqrt{5} \cdot 2\sqrt{4} =$
8. $2\sqrt{4x^3y} \cdot y\sqrt{x^5y^7} =$

Question

Accepted Answer

### Problem 1: Simplify $\boldsymbol{\sqrt{x^2y^{10}}}$
# Explanation:
## Step1: Apply square - root property
For a square root $\sqrt{ab}=\sqrt{a}\cdot\sqrt{b}$ (where $a = x^{2}$ and $b=y^{10}$) and $\sqrt{a^{n}}=a^{\frac{n}{2}}$ for non - negative $a$ and even $n$.
$\sqrt{x^{2}y^{10}}=\sqrt{x^{2}}\cdot\sqrt{y^{10}}$
## Step2: Simplify each square - root
We know that $\sqrt{x^{2}} = x$ (assuming $x\geq0$) and $\sqrt{y^{10}}=y^{5}$ (since $\frac{10}{2}=5$).
So $\sqrt{x^{2}}\cdot\sqrt{y^{10}}=x\cdot y^{5}=xy^{5}$
# Answer:
$xy^{5}$

### Problem 2: Simplify $\boldsymbol{\sqrt{27x^{8}}}$
# Explanation:
## Step1: Factor the radicand
Factor $27$ as $9	imes3$ and $x^{8}$ remains as it is. So $\sqrt{27x^{8}}=\sqrt{9	imes3	imes x^{8}}$
## Step2: Apply square - root property
Using $\sqrt{ab}=\sqrt{a}\cdot\sqrt{b}$, we get $\sqrt{9	imes3	imes x^{8}}=\sqrt{9}\cdot\sqrt{3}\cdot\sqrt{x^{8}}$
## Step3: Simplify each square - root
We know that $\sqrt{9} = 3$, $\sqrt{x^{8}}=x^{4}$ (since $\frac{8}{2}=4$).
So $\sqrt{9}\cdot\sqrt{3}\cdot\sqrt{x^{8}}=3\cdot x^{4}\cdot\sqrt{3}=3x^{4}\sqrt{3}$
# Answer:
$3x^{4}\sqrt{3}$

### Problem 3: Simplify $\boldsymbol{\sqrt{x^{16}}}$
# Explanation:
## Step1: Apply square - root property
For $\sqrt{a^{n}}=a^{\frac{n}{2}}$ (where $a = x$ and $n = 16$).
$\sqrt{x^{16}}=x^{\frac{16}{2}}$
## Step2: Simplify the exponent
$\frac{16}{2}=8$, so $x^{\frac{16}{2}}=x^{8}$
# Answer:
$x^{8}$

### Problem 4: Simplify $\boldsymbol{\sqrt{125b^{15}}}$
# Explanation:
## Step1: Factor the radicand
Factor $125$ as $25	imes5$ and $b^{15}$ as $b^{14}	imes b$. So $\sqrt{125b^{15}}=\sqrt{25	imes5	imes b^{14}	imes b}$
## Step2: Apply square - root property
Using $\sqrt{ab}=\sqrt{a}\cdot\sqrt{b}$, we get $\sqrt{25	imes5	imes b^{14}	imes b}=\sqrt{25}\cdot\sqrt{5}\cdot\sqrt{b^{14}}\cdot\sqrt{b}$
## Step3: Simplify each square - root
We know that $\sqrt{25}=5$, $\sqrt{b^{14}}=b^{7}$ (since $\frac{14}{2}=7$).
So $\sqrt{25}\cdot\sqrt{5}\cdot\sqrt{b^{14}}\cdot\sqrt{b}=5\cdot b^{7}\cdot\sqrt{5b}=5b^{7}\sqrt{5b}$
# Answer:
$5b^{7}\sqrt{5b}$

### Problem 5: Simplify $\boldsymbol{\sqrt{x^{2}y^{4}}\cdot2\sqrt{xy}}$
# Explanation:
## Step1: Simplify $\sqrt{x^{2}y^{4}}$
Using $\sqrt{ab}=\sqrt{a}\cdot\sqrt{b}$ and $\sqrt{a^{n}}=a^{\frac{n}{2}}$, we have $\sqrt{x^{2}y^{4}}=\sqrt{x^{2}}\cdot\sqrt{y^{4}}=x\cdot y^{2}$
## Step2: Multiply the two terms
We have $x\cdot y^{2}\cdot2\sqrt{xy}=2xy^{2}\sqrt{xy}$
## Step3: Combine the radicals (optional, but we can also use exponent rules)
We know that $x = x^{1}$, $y^{2}=y^{2}$, and $\sqrt{xy}=x^{\frac{1}{2}}y^{\frac{1}{2}}$. So $x\cdot x^{\frac{1}{2}}=x^{1 + \frac{1}{2}}=x^{\frac{3}{2}}$ and $y^{2}\cdot y^{\frac{1}{2}}=y^{2+\frac{1}{2}}=y^{\frac{5}{2}}$. Then $2x^{\frac{3}{2}}y^{\frac{5}{2}}=2x\sqrt{x}y^{2}\sqrt{y}=2xy^{2}\sqrt{xy}$ (or we can write it as $2x^{\frac{3}{2}}y^{\frac{5}{2}}$). Another way:
First, use the property $\sqrt{a}\cdot\sqrt{b}=\sqrt{ab}$ for multiplication of radicals. $\sqrt{x^{2}y^{4}}\cdot\sqrt{xy}=\sqrt{x^{2}y^{4}\cdot xy}=\sqrt{x^{3}y^{5}}$
Then $2\sqrt{x^{3}y^{5}}$. Factor $x^{3}=x^{2}\cdot x$ and $y^{5}=y^{4}\cdot y$. So $\sqrt{x^{3}y^{5}}=\sqrt{x^{2}\cdot x\cdot y^{4}\cdot y}=\sqrt{x^{2}}\cdot\sqrt{y^{4}}\cdot\sqrt{xy}=x\cdot y^{2}\cdot\sqrt{xy}$
Then $2	imes x\cdot y^{2}\cdot\sqrt{xy}=2xy^{2}\sqrt{xy}$
# Answer:
$2xy^{2}\sqrt{xy}$ (or $2x^{\frac{3}{2}}y^{\frac{5}{2}}$)

### Problem 6: Simplify $\boldsymbol{\sqrt{9}\cdot\sqrt{32}}$
# Explanation:
## Step1: Simplify $\sqrt{9}$ and $\sqrt{32}$
We know that $\sqrt{9}=3$ and $\sqrt{32}=\sqrt{16	imes2}=\sqrt{16}\cdot\sqrt{2}=4\sqrt{2}$
## Step2: Multiply the two results
$3	imes4\sqrt{2}=12\sqrt{2}$
# Answer:
$12\sqrt{2}$

### Problem 7: Simplify $\boldsymbol{3\sqrt{5}\cdot2\sqrt{4}}$
# Explanation:
## Step1: Simplify $\sqrt{4}$
$\sqrt{4}=2$
## Step2: Multiply the coefficients and the radicals
$3	imes2	imes\sqrt{5}	imes2$ (Wait, no. $3\sqrt{5}\cdot2\sqrt{4}=3	imes2	imes\sqrt{5}	imes\sqrt{4}$)
Since $\sqrt{4}=2$, then $3	imes2	imes\sqrt{5}	imes2$ is wrong. Correctly: $3	imes2	imes\sqrt{5}	imes2$ is incorrect. The correct step is:
$3\sqrt{5}\cdot2\sqrt{4}=3	imes2	imes\sqrt{5}	imes\sqrt{4}$
We know that $\sqrt{4}=2$, so $3	imes2	imes\sqrt{5}	imes2$ is wrong. Wait, $3	imes2 = 6$, $\sqrt{4}=2$, so $6	imes\sqrt{5}	imes2$ is wrong. Wait, no: $a\sqrt{b}\cdot c\sqrt{d}=ac\sqrt{bd}$. So $3\sqrt{5}\cdot2\sqrt{4}=3	imes2	imes\sqrt{5	imes4}=6\sqrt{20}$
Then simplify $\sqrt{20}=\sqrt{4	imes5}=2\sqrt{5}$
So $6	imes2\sqrt{5}=12\sqrt{5}$
Or:
First, simplify $\sqrt{4}=2$. Then $3\sqrt{5}\cdot2	imes2=3\sqrt{5}	imes4 = 12\sqrt{5}$
# Answer:
$12\sqrt{5}$

### Problem 8: Simplify $\boldsymbol{2\sqrt{4x^{3}y}\cdot y\sqrt{x^{5}y^{7}}}$
# Explanation:
## Step1: Use the property of multiplying radicals $a\sqrt{b}\cdot c\sqrt{d}=ac\sqrt{bd}$
Here, $a = 2$, $b = 4x^{3}y$, $c = y$, $d=x^{5}y^{7}$. So $2\sqrt{4x^{3}y}\cdot y\sqrt{x^{5}y^{7}}=2	imes y	imes\sqrt{4x^{3}y\cdot x^{5}y^{7}}$
## Step2: Simplify the radicand
$4x^{3}y\cdot x^{5}y^{7}=4x^{3 + 5}y^{1+7}=4x^{8}y^{8}$
## Step3: Simplify the square - root
$\sqrt{4x^{8}y^{8}}=\sqrt{4}\cdot\sqrt{x^{8}}\cdot\sqrt{y^{8}}=2x^{4}y^{4}$
## Step4: Multiply by the coefficient
$2y	imes2x^{4}y^{4}=4x^{4}y^{5}$
# Answer:
$4x^{4}y^{5}$

Sovi.AI - AI Math Tutor

QUESTION IMAGE

Question

Explanation:

Problem 1: Simplify $\boldsymbol{\sqrt{x^2y^{10}}}$

Step1: Apply square - root property

Step2: Simplify each square - root

Step1: Factor the radicand

Step2: Apply square - root property

Step3: Simplify each square - root

Step1: Apply square - root property

Step2: Simplify the exponent

Answer:

Problem 2: Simplify $\boldsymbol{\sqrt{27x^{8}}}$