name: aden s biscw date: ______ per: ____ unit 4: solving quadratic equations homework 4: pure imaginary numbers ** this is a 2 - page document! ** 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. $i^{64}$ 13. $8i \cdot -9i \cdot 6i$ 14. $(-4i)^3 \cdot 2i$ 15. $(2i)^6 \cdot (\sqrt{6})^2$

Question

Accepted Answer

Let's solve problem 1: $\boldsymbol{\sqrt{-25}}$

# Explanation:
## Step 1: Recall the definition of imaginary unit
We know that $i = \sqrt{-1}$, and for any non - negative real number $a$, $\sqrt{-a}=\sqrt{a}\cdot\sqrt{-1}$.
For the expression $\sqrt{-25}$, we can rewrite it as $\sqrt{25	imes(- 1)}$.

## Step 2: Use the property of square roots
Using the property $\sqrt{ab}=\sqrt{a}\cdot\sqrt{b}$ (where $a = 25$ and $b=-1$), we get $\sqrt{25}	imes\sqrt{-1}$.
Since $\sqrt{25} = 5$ and $\sqrt{-1}=i$, the simplified form is $5i$.

# Answer:
$5i$

Let's solve problem 2: $\boldsymbol{\sqrt{-324}}$

# Explanation:
## Step 1: Rewrite the radicand
We can write $\sqrt{-324}$ as $\sqrt{324	imes(-1)}$.

## Step 2: Simplify the square roots
Using the property $\sqrt{ab}=\sqrt{a}\cdot\sqrt{b}$ (with $a = 324$ and $b = - 1$), we have $\sqrt{324}	imes\sqrt{-1}$.
Since $\sqrt{324}=18$ and $\sqrt{-1}=i$, the simplified form is $18i$.

# Answer:
$18i$

Let's solve problem 3: $\boldsymbol{\sqrt{-15}}$

# Explanation:
## Step 1: Rewrite the radicand
We rewrite $\sqrt{-15}$ as $\sqrt{15	imes(-1)}$.

## Step 2: Simplify the square roots
Using the property $\sqrt{ab}=\sqrt{a}\cdot\sqrt{b}$ (where $a = 15$ and $b=-1$), we get $\sqrt{15}	imes\sqrt{-1}$.
Since $\sqrt{-1}=i$, the simplified form is $i\sqrt{15}$.

# Answer:
$i\sqrt{15}$

Let's solve problem 4: $\boldsymbol{\sqrt{-44}}$

# Explanation:
## Step 1: Factor the radicand
We factor $-44$ as $-1	imes4	imes11$. So, $\sqrt{-44}=\sqrt{-1	imes4	imes11}$.

## Step 2: Use square - root properties
Using $\sqrt{abc}=\sqrt{a}\cdot\sqrt{b}\cdot\sqrt{c}$ (where $a=-1$, $b = 4$, $c = 11$), we have $\sqrt{-1}	imes\sqrt{4}	imes\sqrt{11}$.
Since $\sqrt{-1}=i$ and $\sqrt{4} = 2$, the simplified form is $2i\sqrt{11}$.

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

Let's solve problem 5: $\boldsymbol{\sqrt{-252}}$

# Explanation:
## Step 1: Factor the radicand
Factor $-252$ as $-1	imes36	imes7$. So, $\sqrt{-252}=\sqrt{-1	imes36	imes7}$.

## Step 2: Use square - root properties
Using $\sqrt{abc}=\sqrt{a}\cdot\sqrt{b}\cdot\sqrt{c}$ (where $a=-1$, $b = 36$, $c = 7$), we get $\sqrt{-1}	imes\sqrt{36}	imes\sqrt{7}$.
Since $\sqrt{-1}=i$ and $\sqrt{36}=6$, the simplified form is $6i\sqrt{7}$.

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

Let's solve problem 6: $\boldsymbol{\sqrt{-289}}$

# Explanation:
## Step 1: Rewrite the radicand
We can write $\sqrt{-289}$ as $\sqrt{289	imes(-1)}$.

## Step 2: Simplify the square roots
Using the property $\sqrt{ab}=\sqrt{a}\cdot\sqrt{b}$ (with $a = 289$ and $b=-1$), we have $\sqrt{289}	imes\sqrt{-1}$.
Since $\sqrt{289}=17$ and $\sqrt{-1}=i$, the simplified form is $17i$.

# Answer:
$17i$

Let's solve problem 7: $\boldsymbol{\sqrt{-176}}$

# Explanation:
## Step 1: Factor the radicand
Factor $-176$ as $-1	imes16	imes11$. So, $\sqrt{-176}=\sqrt{-1	imes16	imes11}$.

## Step 2: Use square - root properties
Using $\sqrt{abc}=\sqrt{a}\cdot\sqrt{b}\cdot\sqrt{c}$ (where $a=-1$, $b = 16$, $c = 11$), we get $\sqrt{-1}	imes\sqrt{16}	imes\sqrt{11}$.
Since $\sqrt{-1}=i$ and $\sqrt{16} = 4$, the simplified form is $4i\sqrt{11}$.

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

Let's solve problem 8: $\boldsymbol{\sqrt{-8}\cdot\sqrt{24}}$

# Explanation:
## Step 1: Rewrite the square roots with imaginary unit
First, rewrite $\sqrt{-8}$ as $\sqrt{8}\cdot\sqrt{-1}=2\sqrt{2}i$. And $\sqrt{24}=\sqrt{4	imes6}=2\sqrt{6}$.

## Step 2: Multiply the two expressions
Now, multiply $2\sqrt{2}i$ and $2\sqrt{6}$: $(2\sqrt{2}i)	imes(2\sqrt{6})=2	imes2	imes\sqrt{2	imes6}i$.
Simplify $\sqrt{12}=\sqrt{4	imes3} = 2\sqrt{3}$. Then $4	imes2\sqrt{3}i=8\sqrt{3}i$.

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

Let's solve problem 9: $\boldsymbol{\sqrt{-6}\cdot\sqrt{-12}\cdot\sqrt{-5}}$

# Explanation:
## Step 1: Rewrite each square root
Rewrite $\sqrt{-6}=\sqrt{6}i$, $\sqrt{-12}=\sqrt{12}i = 2\sqrt{3}i$, and $\sqrt{-5}=\sqrt{5}i$.

## Step 2: Multiply the expressions
Multiply them together: $(\sqrt{6}i)	imes(2\sqrt{3}i)	imes(\sqrt{5}i)$.
First, multiply the coefficients and the radicals: $2	imes\sqrt{6	imes3	imes5}	imes(i	imes i	imes i)$.
Simplify $\sqrt{90}=\sqrt{9	imes10}=3\sqrt{10}$. And $i	imes i	imes i=i^{3}=-i$ (since $i^{2}=-1$ and $i^{3}=i^{2}	imes i=-i$).
So, $2	imes3\sqrt{10}	imes(-i)=-6\sqrt{10}i$.

# Answer:
$-6\sqrt{10}i$

Let's solve problem 10: $\boldsymbol{i^{28}}$

# Explanation:
## Step 1: Recall the pattern of powers of $i$
We know that the powers of $i$ have a cyclic pattern: $i^{1}=i$, $i^{2}=-1$, $i^{3}=i^{2}	imes i=-i$, $i^{4}=(i^{2})^{2}=(-1)^{2}=1$, and then the pattern repeats every 4 powers.

## Step 2: Divide the exponent by 4
Divide 28 by 4: $28\div4 = 7$ with a remainder of 0.
When the remainder is 0, $i^{n}$ (where $n$ is a multiple of 4) is equal to 1. So, $i^{28}=(i^{4})^{7}=1^{7}=1$.

# Answer:
$1$

Let's solve problem 11: $\boldsymbol{i^{49}}$

# Explanation:
## Step 1: Find the remainder when dividing the exponent by 4
Divide 49 by 4: $49 = 4	imes12+1$.

## Step 2: Use the pattern of powers of $i$
We know that $i^{4k + r}=i^{r}$, where $k$ is an integer and $r$ is the remainder. Here, $k = 12$ and $r = 1$. So, $i^{49}=i^{4	imes12 + 1}=(i^{4})^{12}	imes i^{1}$.
Since $i^{4}=1$, then $(i^{4})^{12}=1^{12}=1$. And $i^{1}=i$. So, $i^{49}=i$.

# Answer:
$i$

Let's solve problem 12: $\boldsymbol{i^{64}}$

# Explanation:
## Step 1: Divide the exponent by 4
Divide 64 by 4: $64\div4=16$ with a remainder of 0.

## Step 2: Use the pattern of powers of $i$
Since $i^{4}=1$, $i^{64}=(i^{4})^{16}$. And $(i^{4})^{16}=1^{16}=1$.

# Answer:
$1$

Let's solve problem 13: $\boldsymbol{8i\cdot(-9i)\cdot6i}$

# Explanation:
## Step 1: Multiply the coefficients and the $i$ terms separately
First, multiply the coefficients: $8	imes(-9)	imes6=-432$.
Then, multiply the $i$ terms: $i	imes i	imes i=i^{3}$.

## Step 2: Simplify $i^{3}$
We know that $i^{3}=i^{2}	imes i=-1	imes i=-i$.

## Step 3: Combine the results
Multiply the coefficient result with the $i^{3}$ result: $-432	imes(-i)=432i$? Wait, no. Wait, $8	imes(-9)=-72$, $-72	imes6=-432$. And $i	imes i	imes i = i^{3}=-i$. So, $-432	imes(-i)$? Wait, no, $8i	imes(-9i)=-72i^{2}=-72	imes(-1) = 72$. Then $72	imes6i = 432i$? Wait, I made a mistake in the first step. Let's correct it:

## Step 1 (corrected): Multiply two terms first
First, multiply $8i$ and $-9i$: $8i	imes(-9i)=-72i^{2}$. Since $i^{2}=-1$, this is $-72	imes(-1)=72$.

## Step 2 (corrected): Multiply the result by $6i$
Now, multiply 72 by $6i$: $72	imes6i = 432i$. Wait, no, wait: $8i\cdot(-9i)=-72i^{2}=72$ (because $i^{2}=-1$). Then $72\cdot6i=432i$. But wait, let's do it again:

$8i	imes(-9i)=-72i^{2}=72$ (since $i^{2}=-1$). Then $72	imes6i = 432i$. But let's check the sign again. $8	imes(-9)=-72$, $i	imes i = i^{2}=-1$, so $-72	imes(-1)=72$. Then $72	imes6i=432i$. But wait, the original expression is $8i\cdot(-9i)\cdot6i$. So first, $8i	imes(-9i)=-72i^{2}=72$. Then $72	imes6i = 432i$. But wait, let's use the rule of exponents for $i$:

$i	imes i	imes i=i^{3}=-i$. And $8	imes(-9)	imes6=-432$. So $-432	imes i^{3}=-432	imes(-i)=432i$. Yes, that's correct.

# Answer:
$432i$

Let's solve problem 14: $\boldsymbol{(-4i)^{3}\cdot2i}$

# Explanation:
## Step 1: Expand $(-4i)^{3}$
Using the formula $(ab)^{n}=a^{n}b^{n}$, we have $(-4i)^{3}=(-4)^{3}	imes i^{3}=-64	imes(-i)$ (since $i^{3}=-i$). So, $(-4i)^{3}=64i$.

## Step 2: Multiply by $2i$
Now, multiply $64i$ by $2i$: $64i	imes2i = 128i^{2}$.

## Step 3: Simplify $i^{2}$
Since $i^{2}=-1$, $128i^{2}=128	imes(-1)=-128$.

# Answer:
$-128$

Let's solve problem 15: $\boldsymbol{(2i)^{5}\cdot(\sqrt{6})^{2}}$

# Explanation:
## Step 1: Expand $(2i)^{5}$
Using the formula $(ab)^{n}=a^{n}b^{n}$, we have $(2i)^{5}=2^{5}	imes i^{5}=32	imes i^{5}$.
We know that $i^{5}=i^{4}	imes i = 1	imes i=i$ (since $i^{4}=1$). So, $(2i)^{5}=32i$.

## Step 2: Simplify $(\sqrt{6})^{2}$
Using the property $(\sqrt{a})^{2}=a$ (for $a\geq0$), we have $(\sqrt{6})^{2}=6$.

## Step 3: Multiply the two results
Multiply $32i$ and $6$: $32i	imes6 = 192i$.

# Answer:
$192i$

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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 1: Rewrite the radicand

Step 2: Simplify the square roots

Step 1: Rewrite the radicand

Step 2: Simplify the square roots

Answer: