Write the following expression in terms of 𝑖𝑖 and simplify. Show all work.
(√−20)(√−45)


To simplify \((\sqrt{-20})(\sqrt{-45})\) in terms of \(i\), we will follow a step-by-step process.

### Step 1: Express the square roots of negative numbers using \(i\)
Recall that \(\sqrt{-a} = \sqrt{a} \cdot i\), where \(a\) is a positive real number.

For \(\sqrt{-20}\):
\[
\sqrt{-20} = \sqrt{20} \cdot i
\]
Now simplify \(\sqrt{20}\). We know that:
\[
\sqrt{20} = \sqrt{4 \times 5} = \sqrt{4} \times \sqrt{5} = 2\sqrt{5}
\]
Thus:
\[
\sqrt{-20} = 2\sqrt{5} \cdot i
\]

For \(\sqrt{-45}\):
\[
\sqrt{-45} = \sqrt{45} \cdot i
\]
Now simplify \(\sqrt{45}\). We know that:
\[
\sqrt{45} = \sqrt{9 \times 5} = \sqrt{9} \times \sqrt{5} = 3\sqrt{5}
\]
Thus:
\[
\sqrt{-45} = 3\sqrt{5} \cdot i
\]

### Step 2: Multiply the two expressions
Now, multiply the two results:
\[
(\sqrt{-20})(\sqrt{-45}) = (2\sqrt{5} \cdot i)(3\sqrt{5} \cdot i)
\]

First, multiply the constants:
\[
(2\sqrt{5})(3\sqrt{5}) = 6 \times (\sqrt{5})^2 = 6 \times 5 = 30
\]

Next, multiply \(i \cdot i = i^2\). Since \(i^2 = -1\), we get:
\[
i^2 = -1
\]

Thus, the overall expression becomes:
\[
30 \times (-1) = -30
\]

### Final Answer:
\[
(\sqrt{-20})(\sqrt{-45}) = -30
\]

Would you like more details or have any questions?

Here are some related questions for practice:
1. Simplify \((\sqrt{-12})(\sqrt{-48})\) in terms of \(i\).
2. Simplify \((\sqrt{-3})(\sqrt{-27})\).
3. What is \((\sqrt{-16})(\sqrt{-9})\)?
4. How would you express \(\sqrt{-50}\) in terms of \(i\)?
5. Simplify \(\sqrt{-1} \cdot \sqrt{-25}\) and express in terms of \(i\).

**Tip**: Remember that \(i^2 = -1\) is a key property when simplifying complex numbers.

Write the following expression in terms of 𝑖 and simplify. Show all work. (√−20)(√−45)

Learn how to simplify the expression (√−20)(√−45) in terms of 𝑖. This detailed solution walks through converting square roots of negative numbers into complex form and applying properties of imaginary numbers. Ideal for high school students studying complex numbers and algebra.

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