• simplify a square root that is not a perfect square by rewriting it as a product of perfect squares and other factors.$sqrt{20} = sqrt{4 cdot 5} = sqrt{4} cdot sqrt{5} = 2sqrt{5}$• simplify a square root with variables by rewriting the expression to show the perfect square factors. then remove them in the radicand.$sqrt{45a^5} = sqrt{9 cdot 5 cdot a^2 cdot a^2 cdot a} leftarrow $ rewrite $a^5$ as $a^2 cdot a^2 cdot a$.$= sqrt{3 cdot 3} cdot sqrt{5} cdot sqrt{a^2} cdot sqrt{a^2} cdot sqrt{a}$$= 3a^2sqrt{5a}$simplify $sqrt{2} cdot sqrt{12x^3}$.step 1 rewrite the second radical to show the perfect square factors.$sqrt{2} cdot sqrt{12x^3}$$= sqrt{2} cdot sqrt{4 cdot square cdot x^2 cdot square}$$= sqrt{2} cdot sqrt{4} cdot sqrt{3} cdot square cdot square$step 2 simplify the perfect squares.$sqrt{2} cdot sqrt{4} cdot sqrt{3} cdot sqrt{x^2} cdot sqrt{x}$$= sqrt{2} cdot square cdot sqrt{3} cdot square cdot sqrt{x}$step 3 multiply the radicals.$sqrt{2} cdot 2 cdot sqrt{3} cdot x cdot sqrt{x}$$= 2x cdot sqrt{square cdot square cdot square}$$= 2x cdot sqrt{square}$(sqrt{2} cdot sqrt{12x^3} = 2xsqrt{6x})practice simplify each radical expression.$1. 3sqrt{5} cdot 2sqrt{5}$$2. 4sqrt{80}$$3. sqrt{3} cdot sqrt{36}$$4. sqrt{18}$$5. sqrt{63}$$6. 2sqrt{28}$$7. sqrt{25y^8}$$8. 6sqrt{48b^7}$$9. sqrt{8} cdot sqrt{32x^9}$

Question

Accepted Answer

### Practice 1: $ 3\sqrt{5} \cdot 2\sqrt{5} $
# Explanation:
## Step 1: Multiply the coefficients and the radicals
We know that $ a\sqrt{b} \cdot c\sqrt{d}=ac\sqrt{bd} $. So, $ 3\sqrt{5} \cdot 2\sqrt{5}=(3\times2)\times\sqrt{5\times5} $
## Step 2: Simplify the product
$ 3\times2 = 6 $ and $ \sqrt{5\times5}=\sqrt{25} = 5 $. Then we multiply these results: $ 6\times5=30 $
# Answer:
$ 30 $

### Practice 2: $ 4\sqrt{80} $
# Explanation:
## Step 1: Factor 80 into perfect square and other factor
We can write $ 80 = 16\times5 $, where $ 16 $ is a perfect square. So, $ \sqrt{80}=\sqrt{16\times5} $
## Step 2: Use the property $ \sqrt{ab}=\sqrt{a}\cdot\sqrt{b} $
$ \sqrt{16\times5}=\sqrt{16}\cdot\sqrt{5}=4\sqrt{5} $
## Step 3: Multiply by the coefficient 4
$ 4\times4\sqrt{5}=16\sqrt{5} $
# Answer:
$ 16\sqrt{5} $

### Practice 3: $ \sqrt{3} \cdot \sqrt{36} $
# Explanation:
## Step 1: Simplify $ \sqrt{36} $
We know that $ \sqrt{36} = 6 $ since $ 6\times6 = 36 $
## Step 2: Multiply with $ \sqrt{3} $
$ \sqrt{3}\times6 = 6\sqrt{3} $
# Answer:
$ 6\sqrt{3} $

### Practice 4: $ \sqrt{18} $
# Explanation:
## Step 1: Factor 18 into perfect square and other factor
We can write $ 18=9\times2 $, where $ 9 $ is a perfect square. So, $ \sqrt{18}=\sqrt{9\times2} $
## Step 2: Use the property $ \sqrt{ab}=\sqrt{a}\cdot\sqrt{b} $
$ \sqrt{9\times2}=\sqrt{9}\cdot\sqrt{2}=3\sqrt{2} $
# Answer:
$ 3\sqrt{2} $

### Practice 5: $ \sqrt{63} $
# Explanation:
## Step 1: Factor 63 into perfect square and other factor
We can write $ 63 = 9\times7 $, where $ 9 $ is a perfect square. So, $ \sqrt{63}=\sqrt{9\times7} $
## Step 2: Use the property $ \sqrt{ab}=\sqrt{a}\cdot\sqrt{b} $
$ \sqrt{9\times7}=\sqrt{9}\cdot\sqrt{7}=3\sqrt{7} $
# Answer:
$ 3\sqrt{7} $

### Practice 6: $ 2\sqrt{28} $
# Explanation:
## Step 1: Factor 28 into perfect square and other factor
We can write $ 28 = 4\times7 $, where $ 4 $ is a perfect square. So, $ \sqrt{28}=\sqrt{4\times7} $
## Step 2: Use the property $ \sqrt{ab}=\sqrt{a}\cdot\sqrt{b} $
$ \sqrt{4\times7}=\sqrt{4}\cdot\sqrt{7}=2\sqrt{7} $
## Step 3: Multiply by the coefficient 2
$ 2\times2\sqrt{7}=4\sqrt{7} $
# Answer:
$ 4\sqrt{7} $

### Practice 7: $ \sqrt{25y^{8}} $
# Explanation:
## Step 1: Use the property $ \sqrt{ab}=\sqrt{a}\cdot\sqrt{b} $ for $ \sqrt{25y^{8}}=\sqrt{25}\cdot\sqrt{y^{8}} $
## Step 2: Simplify each square root
We know that $ \sqrt{25} = 5 $ and for $ \sqrt{y^{8}} $, using the property $ \sqrt{x^{n}}=x^{\frac{n}{2}} $ (when $ n $ is even), we have $ \sqrt{y^{8}}=y^{\frac{8}{2}}=y^{4} $
## Step 3: Multiply the results
$ 5\times y^{4}=5y^{4} $
# Answer:
$ 5y^{4} $

### Practice 8: $ 6\sqrt{48b^{7}} $
# Explanation:
## Step 1: Factor 48 and $ b^{7} $ into perfect square and other factors
We can write $ 48 = 16\times3 $ (where $ 16 $ is a perfect square) and $ b^{7}=b^{6}\times b $ (where $ b^{6} $ is a perfect square since $ (b^{3})^{2}=b^{6} $). So, $ \sqrt{48b^{7}}=\sqrt{16\times3\times b^{6}\times b} $
## Step 2: Use the property $ \sqrt{ab}=\sqrt{a}\cdot\sqrt{b} $
$ \sqrt{16\times3\times b^{6}\times b}=\sqrt{16}\cdot\sqrt{3}\cdot\sqrt{b^{6}}\cdot\sqrt{b}=4\sqrt{3}\cdot b^{3}\cdot\sqrt{b}=4b^{3}\sqrt{3b} $
## Step 3: Multiply by the coefficient 6
$ 6\times4b^{3}\sqrt{3b}=24b^{3}\sqrt{3b} $
# Answer:
$ 24b^{3}\sqrt{3b} $

### Practice 9: $ \sqrt{8} \cdot \sqrt{32x^{9}} $
# Explanation:
## Step 1: Factor 8 and 32 and $ x^{9} $ into perfect square and other factors
We know that $ 8 = 4\times2 $, $ 32 = 16\times2 $ and $ x^{9}=x^{8}\times x $ (where $ 4,16,x^{8} $ are perfect squares). So, $ \sqrt{8}\cdot\sqrt{32x^{9}}=\sqrt{4\times2}\cdot\sqrt{16\times2\times x^{8}\times x} $
## Step 2: Use the property $ \sqrt{ab}=\sqrt{a}\cdot\sqrt{b} $ for each square root
$ \sqrt{4\times2}=2\sqrt{2} $ and $ \sqrt{16\times2\times x^{8}\times x}=\sqrt{16}\cdot\sqrt{2}\cdot\sqrt{x^{8}}\cdot\sqrt{x}=4\sqrt{2}\cdot x^{4}\cdot\sqrt{x}=4x^{4}\sqrt{2x} $
## Step 3: Multiply the two simplified radicals
$ 2\sqrt{2}\times4x^{4}\sqrt{2x}=(2\times4)x^{4}\times\sqrt{2\times2x} $
## Step 4: Simplify the product of square roots
$ \sqrt{2\times2x}=\sqrt{4x}=2\sqrt{x} $ and $ 2\times4 = 8 $, so $ 8x^{4}\times2\sqrt{x}=16x^{4}\sqrt{x} $? Wait, no, let's correct:

Wait, actually, when multiplying $ \sqrt{a}\cdot\sqrt{b}=\sqrt{ab} $. So $ \sqrt{8}\cdot\sqrt{32x^{9}}=\sqrt{8\times32x^{9}} $

First, calculate $ 8\times32 = 256 $, so $ \sqrt{256x^{9}} $

Now, $ 256 = 16^{2} $ and $ x^{9}=x^{8}\times x=(x^{4})^{2}\times x $

So $ \sqrt{256x^{9}}=\sqrt{256}\cdot\sqrt{x^{8}}\cdot\sqrt{x}=16\times x^{4}\times\sqrt{x}=16x^{4}\sqrt{x} $? Wait, no, $ 8\times32 = 256 $, $ \sqrt{256}=16 $, $ \sqrt{x^{9}}=\sqrt{x^{8}\times x}=x^{4}\sqrt{x} $, so $ \sqrt{8}\cdot\sqrt{32x^{9}}=\sqrt{8\times32x^{9}}=\sqrt{256x^{9}} = 16x^{4}\sqrt{x} $? Wait, but let's do it step by step as per radical multiplication:

$ \sqrt{8}\cdot\sqrt{32x^{9}}=\sqrt{8\times32x^{9}}=\sqrt{256x^{9}} $

Since $ 256 = 16^{2} $ and $ x^{9}=x^{8}\cdot x=(x^{4})^{2}\cdot x $

So $ \sqrt{256x^{9}}=\sqrt{256}\cdot\sqrt{x^{8}}\cdot\sqrt{x}=16\cdot x^{4}\cdot\sqrt{x}=16x^{4}\sqrt{x} $

Wait, but let's check with the initial method:

$ \sqrt{8}=2\sqrt{2} $, $ \sqrt{32x^{9}}=\sqrt{16\times2\times x^{8}\times x}=4x^{4}\sqrt{2x} $

Then $ 2\sqrt{2}\times4x^{4}\sqrt{2x}=8x^{4}\sqrt{4x}=8x^{4}\times2\sqrt{x}=16x^{4}\sqrt{x} $. Yes, that's correct.

Wait, but let's re - do:

$ \sqrt{8}\cdot\sqrt{32x^{9}}=\sqrt{8\times32x^{9}}=\sqrt{256x^{9}} $

$ 256 = 16^{2} $, $ x^{9}=x^{8}\cdot x=(x^{4})^{2}\cdot x $

So $ \sqrt{256x^{9}}=\sqrt{256}\cdot\sqrt{x^{8}}\cdot\sqrt{x}=16\cdot x^{4}\cdot\sqrt{x}=16x^{4}\sqrt{x} $

# Explanation (Correct approach):
## Step 1: Use the property $ \sqrt{a}\cdot\sqrt{b}=\sqrt{ab} $
$ \sqrt{8}\cdot\sqrt{32x^{9}}=\sqrt{8\times32x^{9}} $
## Step 2: Calculate the product inside the square root
$ 8\times32 = 256 $, so we have $ \sqrt{256x^{9}} $
## Step 3: Factor into perfect squares and other factors
$ 256 = 16^{2} $ and $ x^{9}=x^{8}\times x=(x^{4})^{2}\times x $
## Step 4: Simplify the square root
$ \sqrt{256x^{9}}=\sqrt{256}\times\sqrt{x^{8}}\times\sqrt{x}=16\times x^{4}\times\sqrt{x}=16x^{4}\sqrt{x} $
# Answer:
$ 16x^{4}\sqrt{x} $

Sovi.AI - AI Math Tutor

QUESTION IMAGE

Question

Explanation:

Practice 1: \( 3\sqrt{5} \cdot 2\sqrt{5} \)

Step 1: Multiply the coefficients and the radicals

Step 2: Simplify the product

Step 1: Factor 80 into perfect square and other factor

Step 2: Use the property \( \sqrt{ab}=\sqrt{a}\cdot\sqrt{b} \)

Step 3: Multiply by the coefficient 4

Step 1: Simplify \( \sqrt{36} \)

Step 2: Multiply with \( \sqrt{3} \)

Answer:

Practice 2: \( 4\sqrt{80} \)