name: aden fasa date: 1-14-26 per: 2nd hour unit 4: solving quadratic equations homework 4: pure imaginary numbers ** this is a 2 - page document! ** check # 8 directions: simplify the expressions below. 1. $\sqrt{-25}$ 2. $\sqrt{-324}$ 3. $\sqrt{-15}$ 4. $\sqrt{-44}$ 5. $\sqrt{-252}$ 6. $\sqrt{-288}$ 7. $\sqrt{-176}$ 8. $\sqrt{-8}\cdot\sqrt{24}$ 9. $\sqrt{-6}\cdot\sqrt{-12}\cdot\sqrt{-5}$ 10. $i^{28}$ 11. $i^{49}$ 12. $i^{86}$ 13. $8i\cdot -9i\cdot 6i$ 14. $(-4i)^{3}\cdot 2i$ 15. $(2i)^{5}\cdot (i\sqrt{6})^{2}$ $\copyright$ gina wilson (all things algebra$^{\circledr}$, llc), 2015 - 2022

Question

Accepted Answer

Let's solve some of these problems step by step. We'll start with problem 1: $\boldsymbol{\sqrt{-25}}$

# Explanation:
## Step 1: Recall the definition of imaginary unit
We know that $i = \sqrt{-1}$, so we can rewrite $\sqrt{-25}$ as $\sqrt{-1 	imes 25}$.
$\sqrt{-25} = \sqrt{-1 	imes 25}$

## Step 2: Use the property of square roots
Using the property $\sqrt{ab} = \sqrt{a} 	imes \sqrt{b}$ (for $a,b \geq 0$; here we extend it to complex numbers), we get $\sqrt{-1 	imes 25} = \sqrt{-1} 	imes \sqrt{25}$.
$\sqrt{-1} 	imes \sqrt{25} = i 	imes 5$ (since $\sqrt{-1} = i$ and $\sqrt{25} = 5$)

## Step 3: Simplify the expression
Multiplying $i$ and $5$ gives $5i$.
$i 	imes 5 = 5i$

# Answer:
$5i$

Now problem 2: $\boldsymbol{\sqrt{-324}}$

# Explanation:
## Step 1: Rewrite the radicand
Rewrite $\sqrt{-324}$ as $\sqrt{-1 	imes 324}$.
$\sqrt{-324} = \sqrt{-1 	imes 324}$

## Step 2: Apply the square root property
Using $\sqrt{ab} = \sqrt{a} 	imes \sqrt{b}$, we have $\sqrt{-1 	imes 324} = \sqrt{-1} 	imes \sqrt{324}$.
$\sqrt{-1} 	imes \sqrt{324} = i 	imes 18$ (since $\sqrt{-1} = i$ and $\sqrt{324} = 18$)

## Step 3: Simplify
Multiplying $i$ and $18$ gives $18i$.
$i 	imes 18 = 18i$

# Answer:
$18i$

Problem 3: $\boldsymbol{\sqrt{-15}}$

# Explanation:
## Step 1: Rewrite the radicand
Rewrite $\sqrt{-15}$ as $\sqrt{-1 	imes 15}$.
$\sqrt{-15} = \sqrt{-1 	imes 15}$

## Step 2: Apply the square root property
Using $\sqrt{ab} = \sqrt{a} 	imes \sqrt{b}$, we get $\sqrt{-1 	imes 15} = \sqrt{-1} 	imes \sqrt{15}$.
$\sqrt{-1} 	imes \sqrt{15} = i 	imes \sqrt{15}$ (since $\sqrt{-1} = i$)

## Step 3: Simplify
This gives $i\sqrt{15}$.
$i 	imes \sqrt{15} = i\sqrt{15}$

# Answer:
$i\sqrt{15}$

Problem 4: $\boldsymbol{\sqrt{-44}}$

# Explanation:
## Step 1: Rewrite the radicand
Rewrite $\sqrt{-44}$ as $\sqrt{-1 	imes 44}$.
$\sqrt{-44} = \sqrt{-1 	imes 44}$

## Step 2: Factor 44
Factor $44$ as $4 	imes 11$, so $\sqrt{-1 	imes 44} = \sqrt{-1 	imes 4 	imes 11}$.

## Step 3: Apply the square root property
Using $\sqrt{abc} = \sqrt{a} 	imes \sqrt{b} 	imes \sqrt{c}$, we get $\sqrt{-1 	imes 4 	imes 11} = \sqrt{-1} 	imes \sqrt{4} 	imes \sqrt{11}$.
$\sqrt{-1} 	imes \sqrt{4} 	imes \sqrt{11} = i 	imes 2 	imes \sqrt{11}$ (since $\sqrt{-1} = i$, $\sqrt{4} = 2$)

## Step 4: Simplify
Multiplying $i$, $2$, and $\sqrt{11}$ gives $2i\sqrt{11}$.
$i 	imes 2 	imes \sqrt{11} = 2i\sqrt{11}$

# Answer:
$2i\sqrt{11}$

Problem 5: $\boldsymbol{\sqrt{-252}}$

# Explanation:
## Step 1: Rewrite the radicand
Rewrite $\sqrt{-252}$ as $\sqrt{-1 	imes 252}$.
$\sqrt{-252} = \sqrt{-1 	imes 252}$

## Step 2: Factor 252
Factor $252$ as $36 	imes 7$, so $\sqrt{-1 	imes 252} = \sqrt{-1 	imes 36 	imes 7}$.

## Step 3: Apply the square root property
Using $\sqrt{abc} = \sqrt{a} 	imes \sqrt{b} 	imes \sqrt{c}$, we get $\sqrt{-1 	imes 36 	imes 7} = \sqrt{-1} 	imes \sqrt{36} 	imes \sqrt{7}$.
$\sqrt{-1} 	imes \sqrt{36} 	imes \sqrt{7} = i 	imes 6 	imes \sqrt{7}$ (since $\sqrt{-1} = i$, $\sqrt{36} = 6$)

## Step 4: Simplify
Multiplying $i$, $6$, and $\sqrt{7}$ gives $6i\sqrt{7}$.
$i 	imes 6 	imes \sqrt{7} = 6i\sqrt{7}$

# Answer:
$6i\sqrt{7}$

Problem 6: $\boldsymbol{\sqrt{-288}}$

# Explanation:
## Step 1: Rewrite the radicand
Rewrite $\sqrt{-288}$ as $\sqrt{-1 	imes 288}$.
$\sqrt{-288} = \sqrt{-1 	imes 288}$

## Step 2: Factor 288
Factor $288$ as $144 	imes 2$, so $\sqrt{-1 	imes 288} = \sqrt{-1 	imes 144 	imes 2}$.

## Step 3: Apply the square root property
Using $\sqrt{abc} = \sqrt{a} 	imes \sqrt{b} 	imes \sqrt{c}$, we get $\sqrt{-1 	imes 144 	imes 2} = \sqrt{-1} 	imes \sqrt{144} 	imes \sqrt{2}$.
$\sqrt{-1} 	imes \sqrt{144} 	imes \sqrt{2} = i 	imes 12 	imes \sqrt{2}$ (since $\sqrt{-1} = i$, $\sqrt{144} = 12$)

## Step 4: Simplify
Multiplying $i$, $12$, and $\sqrt{2}$ gives $12i\sqrt{2}$.
$i 	imes 12 	imes \sqrt{2} = 12i\sqrt{2}$

# Answer:
$12i\sqrt{2}$

Problem 7: $\boldsymbol{\sqrt{-176}}$

# Explanation:
## Step 1: Rewrite the radicand
Rewrite $\sqrt{-176}$ as $\sqrt{-1 	imes 176}$.
$\sqrt{-176} = \sqrt{-1 	imes 176}$

## Step 2: Factor 176
Factor $176$ as $16 	imes 11$, so $\sqrt{-1 	imes 176} = \sqrt{-1 	imes 16 	imes 11}$.

## Step 3: Apply the square root property
Using $\sqrt{abc} = \sqrt{a} 	imes \sqrt{b} 	imes \sqrt{c}$, we get $\sqrt{-1 	imes 16 	imes 11} = \sqrt{-1} 	imes \sqrt{16} 	imes \sqrt{11}$.
$\sqrt{-1} 	imes \sqrt{16} 	imes \sqrt{11} = i 	imes 4 	imes \sqrt{11}$ (since $\sqrt{-1} = i$, $\sqrt{16} = 4$)

## Step 4: Simplify
Multiplying $i$, $4$, and $\sqrt{11}$ gives $4i\sqrt{11}$.
$i 	imes 4 	imes \sqrt{11} = 4i\sqrt{11}$

# Answer:
$4i\sqrt{11}$

Problem 8: $\boldsymbol{\sqrt{-8} \cdot \sqrt{24}}$

# Explanation:
## Step 1: Rewrite $\sqrt{-8}$
Rewrite $\sqrt{-8}$ as $\sqrt{-1 	imes 8} = \sqrt{-1} 	imes \sqrt{8} = i 	imes 2\sqrt{2}$ (since $\sqrt{8} = 2\sqrt{2}$)

## Step 2: Simplify $\sqrt{24}$
Simplify $\sqrt{24}$ as $\sqrt{4 	imes 6} = \sqrt{4} 	imes \sqrt{6} = 2\sqrt{6}$

## Step 3: Multiply the two expressions
Now multiply $i 	imes 2\sqrt{2}$ and $2\sqrt{6}$:
$(i 	imes 2\sqrt{2}) 	imes (2\sqrt{6}) = i 	imes 2 	imes 2 	imes \sqrt{2} 	imes \sqrt{6}$

## Step 4: Simplify the product of square roots
$\sqrt{2} 	imes \sqrt{6} = \sqrt{12} = \sqrt{4 	imes 3} = 2\sqrt{3}$

## Step 5: Multiply the coefficients and the imaginary unit
$i 	imes 2 	imes 2 	imes 2\sqrt{3} = i 	imes 8\sqrt{3}$? Wait, no, let's check again. Wait, $2 	imes 2 = 4$, and $\sqrt{2} 	imes \sqrt{6} = \sqrt{12} = 2\sqrt{3}$, so $4 	imes 2\sqrt{3} = 8\sqrt{3}$? Wait, no, that's not right. Wait, let's do it again.

Wait, $\sqrt{-8} = i\sqrt{8} = i 	imes 2\sqrt{2}$ (correct, since $\sqrt{8} = 2\sqrt{2}$)

$\sqrt{24} = \sqrt{4 	imes 6} = 2\sqrt{6}$ (correct)

Now multiply them: $(i 	imes 2\sqrt{2}) 	imes (2\sqrt{6}) = i 	imes (2\sqrt{2} 	imes 2\sqrt{6}) = i 	imes 4 	imes \sqrt{12}$ (since $\sqrt{2} 	imes \sqrt{6} = \sqrt{12}$)

$\sqrt{12} = 2\sqrt{3}$, so $4 	imes 2\sqrt{3} = 8\sqrt{3}$? Wait, no, $2\sqrt{2} 	imes 2\sqrt{6} = 4 	imes \sqrt{12} = 4 	imes 2\sqrt{3} = 8\sqrt{3}$? But that seems off. Wait, maybe a better way:

$\sqrt{-8} \cdot \sqrt{24} = \sqrt{-8 	imes 24}$ (using the property $\sqrt{a} \cdot \sqrt{b} = \sqrt{ab}$ for $a,b \geq 0$; here we extend it to complex numbers)

So $\sqrt{-8 	imes 24} = \sqrt{-192}$

Now simplify $\sqrt{-192}$:

$\sqrt{-192} = \sqrt{-1 	imes 64 	imes 3} = \sqrt{-1} 	imes \sqrt{64} 	imes \sqrt{3} = i 	imes 8 	imes \sqrt{3} = 8i\sqrt{3}$

Ah, that's a better approach. So using $\sqrt{a} \cdot \sqrt{b} = \sqrt{ab}$ (for complex numbers, we can use this property as long as we handle the negative sign with $i$). So $\sqrt{-8} \cdot \sqrt{24} = \sqrt{-8 	imes 24} = \sqrt{-192}$

Then $\sqrt{-192} = \sqrt{-1 	imes 64 	imes 3} = \sqrt{-1} 	imes \sqrt{64} 	imes \sqrt{3} = i 	imes 8 	imes \sqrt{3} = 8i\sqrt{3}$

# Answer:
$8i\sqrt{3}$

Problem 9: $\boldsymbol{\sqrt{-6} \cdot \sqrt{-12} \cdot \sqrt{-5}}$

# Explanation:
## Step 1: Rewrite each square root
$\sqrt{-6} = i\sqrt{6}$, $\sqrt{-12} = i\sqrt{12}$, $\sqrt{-5} = i\sqrt{5}$

## Step 2: Multiply the imaginary units and the square roots
Multiply them together: $(i\sqrt{6}) \cdot (i\sqrt{12}) \cdot (i\sqrt{5})$

## Step 3: Multiply the imaginary units
$i \cdot i \cdot i = i^3$

## Step 4: Multiply the square roots
$\sqrt{6} \cdot \sqrt{12} \cdot \sqrt{5} = \sqrt{6 	imes 12 	imes 5} = \sqrt{360}$

## Step 5: Simplify $i^3$ and $\sqrt{360}$
We know that $i^3 = -i$ (since $i^2 = -1$, so $i^3 = i^2 \cdot i = -1 \cdot i = -i$)

$\sqrt{360} = \sqrt{36 	imes 10} = 6\sqrt{10}$

## Step 6: Multiply the results
$-i \cdot 6\sqrt{10} = -6i\sqrt{10}$? Wait, but the handwritten answer is $6i\sqrt{10}$. Did we make a mistake?

Wait, let's check again. Wait, $\sqrt{-6} \cdot \sqrt{-12} \cdot \sqrt{-5} = (i\sqrt{6})(i\sqrt{12})(i\sqrt{5}) = i^3 \cdot \sqrt{6 	imes 12 	imes 5}$

$i^3 = -i$, and $\sqrt{6 	imes 12 	imes 5} = \sqrt{360} = 6\sqrt{10}$, so $-i 	imes 6\sqrt{10} = -6i\sqrt{10}$. But the handwritten answer is $6i\sqrt{10}$. Maybe there's a mistake in the handwritten answer or in our approach. Wait, maybe we should use the property $\sqrt{-a} \cdot \sqrt{-b} = -\sqrt{ab}$ for $a,b > 0$. Let's recall that:

$\sqrt{-a} \cdot \sqrt{-b} = \sqrt{-1 	imes a} \cdot \sqrt{-1 	imes b} = i\sqrt{a} \cdot i\sqrt{b} = i^2 \sqrt{ab} = - \sqrt{ab}$

So let's apply that:

First, multiply $\sqrt{-6}$ and $\sqrt{-12}$:

$\sqrt{-6} \cdot \sqrt{-12} = - \sqrt{6 	imes 12} = - \sqrt{72} = - 6\sqrt{2}$

Then multiply that result by $\sqrt{-5}$:

$- 6\sqrt{2} \cdot \sqrt{-5} = - 6\sqrt{2} \cdot i\sqrt{5} = - 6i\sqrt{10}$

Wait, so that's $-6i\sqrt{10}$, but the handwritten answer is $6i\sqrt{10}$. Maybe the handwritten answer is incorrect, or maybe we misread the problem. Let's check the problem again: $\sqrt{-6} \cdot \sqrt{-12} \cdot \sqrt{-5}$. Yes, that's what's written. So according to the rules of complex numbers, the correct answer should be $-6i\sqrt{10}$. But let's see the handwritten work: $i\sqrt{6} \cdot i\sqrt{12} \cdot i\sqrt{5} = i^3 \sqrt{6 	imes 12 	imes 5} = -i 	imes 6\sqrt{10} = -6i\sqrt{10}$. So the handwritten answer has a sign error.

Problem 10: $\boldsymbol{i^{28}}$

# Explanation:
## Step 1: Recall the powers of $i$
We know that the powers of $i$ cycle every 4:
$i^1 = i$
$i^2 = -1$
$i^3 = -i$
$i^4 = 1$
Then it repeats: $i^5 = i$, $i^6 = -1$, etc.

## Step 2: Divide the exponent by 4
To find $i^{28}$, divide 28 by 4: $28 \div 4 = 7$ with a remainder of 0.

## Step 3: Determine the power
Since the remainder is 0

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QUESTION IMAGE

Question

Explanation:

Step 1: Recall the definition of imaginary unit

Step 2: Use the property of square roots

Step 3: Simplify the expression

Step 1: Rewrite the radicand

Step 2: Apply the square root property

Step 3: Simplify

Step 1: Rewrite the radicand

Step 2: Apply the square root property

Step 3: Simplify

Answer: