Question

algebra ii practice n.cn.a.2: square roots of negative numbers www.jmap.org
name: kaleb rodgers

1. simplify. \\(\sqrt{-9}\\)
2. express \\(\sqrt{-8}\\) in \\(i\\) notation.

a \\(8i\\) b \\(2i\sqrt{-2}\\)
c \\(\sqrt{-8}i\\) d \\(2i\sqrt{2}\\)

1. express \\(\sqrt{-80}\\) in \\(i\\) notation.

a \\(4i\sqrt{-5}\\) b \\(\sqrt{-80}i\\)
c \\(4i\sqrt{5}\\) d \\(80i\\)

1. express \\(\sqrt{-75}\\) in \\(i\\) notation.

a \\(75i\\) b \\(5i\sqrt{-3}\\)
c \\(\sqrt{-75}i\\) d \\(5i\sqrt{3}\\)

1. express \\(\sqrt{-72}\\) in \\(i\\) notation.

a \\(6i\sqrt{-2}\\) b \\(\sqrt{-72}i\\)
c \\(6i\sqrt{2}\\) d \\(72i\\)

1. express \\(\sqrt{-20}\\) in \\(i\\) notation.

a \\(20i\\) b \\(2i\sqrt{5}\\)
c \\(2i\sqrt{-5}\\) d \\(\sqrt{-20}i\\)

1. express \\(\sqrt{-27}\\) in \\(i\\) notation.

a \\(\sqrt{-27}i\\) b \\(3i\sqrt{-3}\\)
c \\(3i\sqrt{3}\\) d \\(27i\\)

1. express \\(\sqrt{-12}\\) in \\(i\\) notation.

a \\(2i\sqrt{-3}\\) b \\(\sqrt{-12}i\\)
c \\(12i\\) d \\(2i\sqrt{3}\\)

1. express \\(\sqrt{-125}\\) in \\(i\\) notation.

a \\(125i\\) b \\(5i\sqrt{-5}\\)
c \\(5i\sqrt{5}\\) d \\(\sqrt{-125}i\\)

1. express \\(\sqrt{-180}\\) in \\(i\\) notation.

a \\(180i\\) b \\(6i\sqrt{-5}\\)
c \\(6\sqrt{5}\\) d \\(\sqrt{-180}i\\)

Explanation:

Step1: Factor out -1 from radicand

For $\sqrt{-n}$, rewrite as $\sqrt{-1 \times n} = \sqrt{-1} \times \sqrt{n}$, where $i = \sqrt{-1}$.

Step2: Simplify positive radical

Reduce $\sqrt{n}$ to simplest radical form by factoring out perfect squares.

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Problem 1

Step1: Split into $\sqrt{-1}\sqrt{9}$

$\sqrt{-9} = \sqrt{-1} \times \sqrt{9}$

Step2: Substitute $i$ and simplify

$\sqrt{-9} = i \times 3 = 3i$

Problem 2

Step1: Split into $\sqrt{-1}\sqrt{8}$

$\sqrt{-8} = \sqrt{-1} \times \sqrt{8}$

Step2: Simplify $\sqrt{8}$ and substitute $i$

$\sqrt{8} = \sqrt{4 \times 2} = 2\sqrt{2}$, so $\sqrt{-8} = 2i\sqrt{2}$

Problem 3

Step1: Split into $\sqrt{-1}\sqrt{80}$

$\sqrt{-80} = \sqrt{-1} \times \sqrt{80}$

Step2: Simplify $\sqrt{80}$ and substitute $i$

$\sqrt{80} = \sqrt{16 \times 5} = 4\sqrt{5}$, so $\sqrt{-80} = 4i\sqrt{5}$

Problem 4

Step1: Split into $\sqrt{-1}\sqrt{75}$

$\sqrt{-75} = \sqrt{-1} \times \sqrt{75}$

Step2: Simplify $\sqrt{75}$ and substitute $i$

$\sqrt{75} = \sqrt{25 \times 3} = 5\sqrt{3}$, so $\sqrt{-75} = 5i\sqrt{3}$

Problem 5

Step1: Split into $\sqrt{-1}\sqrt{72}$

$\sqrt{-72} = \sqrt{-1} \times \sqrt{72}$

Step2: Simplify $\sqrt{72}$ and substitute $i$

$\sqrt{72} = \sqrt{36 \times 2} = 6\sqrt{2}$, so $\sqrt{-72} = 6i\sqrt{2}$

Problem 6

Step1: Split into $\sqrt{-1}\sqrt{20}$

$\sqrt{-20} = \sqrt{-1} \times \sqrt{20}$

Step2: Simplify $\sqrt{20}$ and substitute $i$

$\sqrt{20} = \sqrt{4 \times 5} = 2\sqrt{5}$, so $\sqrt{-20} = 2i\sqrt{5}$

Problem 7

Step1: Split into $\sqrt{-1}\sqrt{27}$

$\sqrt{-27} = \sqrt{-1} \times \sqrt{27}$

Step2: Simplify $\sqrt{27}$ and substitute $i$

$\sqrt{27} = \sqrt{9 \times 3} = 3\sqrt{3}$, so $\sqrt{-27} = 3i\sqrt{3}$

Problem 8

Step1: Split into $\sqrt{-1}\sqrt{12}$

$\sqrt{-12} = \sqrt{-1} \times \sqrt{12}$

Step2: Simplify $\sqrt{12}$ and substitute $i$

$\sqrt{12} = \sqrt{4 \times 3} = 2\sqrt{3}$, so $\sqrt{-12} = 2i\sqrt{3}$

Problem 9

Step1: Split into $\sqrt{-1}\sqrt{125}$

$\sqrt{-125} = \sqrt{-1} \times \sqrt{125}$

Step2: Simplify $\sqrt{125}$ and substitute $i$

$\sqrt{125} = \sqrt{25 \times 5} = 5\sqrt{5}$, so $\sqrt{-125} = 5i\sqrt{5}$

Problem 10

Step1: Split into $\sqrt{-1}\sqrt{180}$

$\sqrt{-180} = \sqrt{-1} \times \sqrt{180}$

Step2: Simplify $\sqrt{180}$ and substitute $i$

$\sqrt{180} = \sqrt{36 \times 5} = 6\sqrt{5}$, so $\sqrt{-180} = 6i\sqrt{5}$

Answer:

1. $3i$
2. D. $2i\sqrt{2}$
3. $4i\sqrt{5}$
4. D. $5i\sqrt{3}$
5. C. $6i\sqrt{2}$
6. B. $2i\sqrt{5}$
7. C. $3i\sqrt{3}$
8. D. $2i\sqrt{3}$
9. C. $5i\sqrt{5}$
10. $6i\sqrt{5}$