Question

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. $i^{86}$
13. $8i \cdot -9i \cdot 6i$
14. $(-4i)^3 \cdot 2i$
15. $(2i)^5 \cdot (i\sqrt{6})^2$

© gina wilson (all things algebra®, llc), 2013 - 2022

Explanation:

Step1: Simplify \(\sqrt{-25}\)

Recall that \(\sqrt{-a}=\sqrt{a}\cdot i\) for \(a>0\), where \(i\) is the imaginary unit with \(i^2 = - 1\). So, \(\sqrt{-25}=\sqrt{25}\cdot i\). Since \(\sqrt{25} = 5\), we have \(\sqrt{-25}=5i\).

Step2: Simplify \(\sqrt{-324}\)

Using the rule \(\sqrt{-a}=\sqrt{a}\cdot i\), we get \(\sqrt{-324}=\sqrt{324}\cdot i\). And \(\sqrt{324}=18\), so \(\sqrt{-324}=18i\).

Step3: Simplify \(\sqrt{-15}\)

Applying \(\sqrt{-a}=\sqrt{a}\cdot i\), we have \(\sqrt{-15}=\sqrt{15}\cdot i=\sqrt{15}i\).

Step4: Simplify \(\sqrt{-44}\)

First, factor \(44 = 4\times11\). Then \(\sqrt{-44}=\sqrt{4\times11\times(- 1)}=\sqrt{4}\cdot\sqrt{11}\cdot\sqrt{-1}\). Since \(\sqrt{4} = 2\) and \(\sqrt{-1}=i\), we get \(2\sqrt{11}i\).

Step5: Simplify \(\sqrt{-252}\)

Factor \(252=36\times7\). So, \(\sqrt{-252}=\sqrt{36\times7\times(-1)}=\sqrt{36}\cdot\sqrt{7}\cdot\sqrt{-1}\). As \(\sqrt{36} = 6\) and \(\sqrt{-1}=i\), we have \(6\sqrt{7}i\).

Step6: Simplify \(\sqrt{-288}\)

Factor \(288 = 144\times2\). Then \(\sqrt{-288}=\sqrt{144\times2\times(-1)}=\sqrt{144}\cdot\sqrt{2}\cdot\sqrt{-1}\). Since \(\sqrt{144}=12\) and \(\sqrt{-1}=i\), we get \(12\sqrt{2}i\).

Step7: Simplify \(\sqrt{-176}\)

Factor \(176 = 16\times11\). So, \(\sqrt{-176}=\sqrt{16\times11\times(-1)}=\sqrt{16}\cdot\sqrt{11}\cdot\sqrt{-1}\). As \(\sqrt{16} = 4\) and \(\sqrt{-1}=i\), we have \(4\sqrt{11}i\).

Step8: Simplify \(\sqrt{-8}\cdot\sqrt{24}\)

First, \(\sqrt{-8}=\sqrt{8}\cdot i = 2\sqrt{2}i\) and \(\sqrt{24}=2\sqrt{6}\). Then multiply them: \((2\sqrt{2}i)\cdot(2\sqrt{6})=4\sqrt{12}i\). Simplify \(\sqrt{12}=2\sqrt{3}\), so \(4\times2\sqrt{3}i = 8\sqrt{3}i\). Wait, another way: \(\sqrt{-8}\cdot\sqrt{24}=\sqrt{-8\times24}=\sqrt{-192}\). Factor \(192 = 64\times3\), so \(\sqrt{-192}=\sqrt{64\times3}\cdot i=8\sqrt{3}i\).

Step9: Simplify \(\sqrt{-6}\cdot\sqrt{-12}\cdot\sqrt{-5}\)

We know \(\sqrt{-a}=i\sqrt{a}\), so \(\sqrt{-6}=i\sqrt{6}\), \(\sqrt{-12}=i\sqrt{12}=i\cdot2\sqrt{3}\), \(\sqrt{-5}=i\sqrt{5}\). Multiply them: \((i\sqrt{6})\cdot(2i\sqrt{3})\cdot(i\sqrt{5})\). First, multiply the coefficients and the \(i\) terms: \(2\times i\times i\times i\times\sqrt{6\times3\times5}\). \(i\times i\times i=i^3=-i\), and \(\sqrt{90}=3\sqrt{10}\). So \(2\times(-i)\times3\sqrt{10}=-6\sqrt{10}i\). Wait, or using \(\sqrt{-a}\cdot\sqrt{-b}=\sqrt{ab}\cdot i^2=-\sqrt{ab}\) (since \(i^2=-1\)). So \(\sqrt{-6}\cdot\sqrt{-12}=-\sqrt{72}\), then multiply by \(\sqrt{-5}\): \(-\sqrt{72}\cdot\sqrt{-5}=-\sqrt{72}\cdot i\sqrt{5}=-i\sqrt{360}=-i\cdot6\sqrt{10}=-6\sqrt{10}i\).

Step10: Simplify \(i^{28}\)

Recall that \(i\) has a cycle of 4: \(i^1 = i\), \(i^2=-1\), \(i^3=-i\), \(i^4 = 1\), and then repeats. So divide the exponent by 4: \(28\div4 = 7\) with remainder 0. So \(i^{28}=(i^4)^7=1^7 = 1\).

Step11: Simplify \(i^{49}\)

Divide 49 by 4: \(49 = 4\times12+1\). So \(i^{49}=(i^4)^{12}\cdot i^1=1^{12}\cdot i = i\).

Step12: Simplify \(i^{86}\)

Divide 86 by 4: \(86 = 4\times21+2\). So \(i^{86}=(i^4)^{21}\cdot i^2=1^{21}\cdot(-1)=-1\).

Step13: Simplify \(8i\cdot - 9i\cdot6i\)

First, multiply the coefficients: \(8\times(-9)\times6=-432\). Multiply the \(i\) terms: \(i\cdot i\cdot i=i^3=-i\). So the result is \(-432\times(-i)=432i\)? Wait, no: \(8i\times(-9i)=-72i^2=-72\times(-1)=72\). Then \(72\times6i = 432i\). Wait, \(i^2=-1\), so \(8i\times(-9i)=8\times(-9)\times i^2=-72\times(-1)=72\). Then \(72\times6i = 432i\).

Step14: Simplify \((-4i)^3\cdot2i\)

First, expand \((-4i)^3=(-4)^3\cdot i^3=-64\cdot(-i)=64i\) (since \(i^3=-i\)). Then multiply by \(2i\): \(64i\times2i = 128i^2=128\times(-1)=-128\).

Step15: Simplify \((2i)^5\cdot(i\sqrt{6})^2\)

Expand \((2i)^5=2^5\cdot i^5=32\cdot i^4\cdot i=32\times1\times i = 32i\). Expand \((i\sqrt{6})^2=i^2\times6=-6\). Then multiply them: \(32i\times(-6)=-192i\).

Answer:

1. \(5i\)
2. \(18i\)
3. \(\sqrt{15}i\)
4. \(2\sqrt{11}i\)
5. \(6\sqrt{7}i\)
6. \(12\sqrt{2}i\)
7. \(4\sqrt{11}i\)
8. \(8\sqrt{3}i\)
9. \(-6\sqrt{10}i\)
10. \(1\)
11. \(i\)
12. \(-1\)
13. \(432i\)
14. \(-128\)
15. \(-192i\)