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Algebra II : Mathematical Relationships and Basic Graphs

Study concepts, example questions & explanations for Algebra II

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Example Questions

Example Question #11 : Sat Subject Test In Math I

Simplify:

(2+5i)(2-5i)

Possible Answers:

4-10i

16-5i

Correct answer:

Explanation:

Start by using FOIL. Which means to multiply the first terms together then the outer terms followed by the inner terms and lastly, the last terms.

(2+5i)(2-5i)=4-10i+10i-25i^2

Remember that $i=\sqrt{-1}$ , so i^2=-1 .

Substitute in for i^2 .

4-10i+10i-25i^2=4-25(-1)=4+25=29

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Example Question #31 : Imaginary Numbers & Complex Functions

Simplify:

$\frac{5-2i}{4-i}$

Possible Answers:

6+3i

18-9i

$\frac{22-3i}{17}$

$\frac{5i+6}{6}$

Correct answer:

$\frac{22-3i}{17}$

Explanation:

To get rid of the fraction, multiply the numerator and denominator by the conjugate of the denominator.

$\frac{5-2i}{4-i}\left(\frac{4+i}{4+i}\right)$

Now, multiply and simplify.

$\frac{5-2i}{4-i}\left(\frac{4+i}{4+i}\right)=\frac{20+5i-8i-2i^2}{16-i^2}$

Remember that i^2=-1

$\frac{20+5i-8i-2i^2}{16-i^2}=\frac{20-3i-2(-1)}{16-(-1)}=\frac{22-3i}{17}$

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Example Question #1 : Complex Conjugates

Simplify:

$\frac{8-3i}{2-i}$

Possible Answers:

10-7i

$\frac{19+2i}{5}$

3i+6

$\frac{7+2i}{3}$

Correct answer:

$\frac{19+2i}{5}$

Explanation:

To get rid of the fraction, multiply the numerator and denominator by the conjugate of the denominator.

$\frac{8-3i}{2-i}\left(\frac{2+i}{2+i}\right)$

Now, multiply and simplify.

$\frac{8-3i}{2-i}\left(\frac{2+i}{2+i}\right)=\frac{16+8i-6i-3i^2}{4-i^2}$

Remember that i^2=-1

$\frac{16+8i-6i-3i^2}{4-i^2}=\frac{16+2i-3(-1)}{4-(-1)}=\frac{19+2i}{5}$

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Example Question #15 : Number Theory

Write in standard form: $\frac{3}{2+4i}$

Possible Answers:

$\frac{6-12i}{20}$

6-12i

$\frac{2+4i}{3}$

None of the other answers

$\frac{3}{10}-\frac{3i}{5}$

Correct answer:

$\frac{3}{10}-\frac{3i}{5}$

Explanation:

$\frac{3}{2+4i}$

Multiply by the conjugate:

$\frac{3}{2+4i}*\frac{2-4i}{2-4i}$

$\frac{3(2-4i)}{4-16i^2}$

$\frac{6-12i}{4-16(-1)}$

$\frac{6-12i}{20}=\frac{6}{20}-\frac{12i}{20}=\mathbf{\frac{3}{10}-\frac{3i}{5}}$

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Example Question #12 : Imaginary Numbers

Write in standard form: $\frac{12i}{2+4i}$

Possible Answers:

$\frac{6i}{5}-5$

$\frac{6i}{5}-\frac{12}{5}$

$\frac{6i}{5}+\frac{12}{5}$

None of these.

$\frac{6}{5}+\frac{12i}{5}$

Correct answer:

$\frac{6i}{5}+\frac{12}{5}$

Explanation:

$\frac{12i}{2+4i}$

Multiply by the conjugate:

$\frac{12i}{2+4i}*\frac{2-4i}{2-4i}$

$\frac{12i(2-4i)}{4-16i^2}$

Combine:

$\frac{24i-48i^2}{4-16i^2}$

Simplify:

$\frac{24i+48}{20}$

$\mathbf{\frac{6i}{5}+\frac{12}{5}}$

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Example Question #13 : Imaginary Numbers

Simplify:

$\sqrt{-64}$

Possible Answers:

None of the above

64i

$\pm8i$

Correct answer:

$\pm8i$

Explanation:

To solve a radical that has a negative sign under it we need to factor it first.

$\sqrt{-64} = \sqrt{(-1)*64}$

$= \sqrt{(-1)}*\sqrt{64}$

Recall that $\sqrt{-1}=i$ . Using this fact we get the following.

$= i*\pm8 = \pm8i$

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Example Question #1971 : Mathematical Relationships And Basic Graphs

Simplify:

$\sqrt{-242}$

Possible Answers:

$i\sqrt{242}$

None of the above

$\pm11i\sqrt{2}$

11i

Correct answer:

$\pm11i\sqrt{2}$

Explanation:

To simplify a radical that has a negative sign under it we need to factor it first.

After factoring it recall that $\sqrt{-1}=i$ .

Therefore we get the following.

$\sqrt{-242} = \sqrt{(-1)*(2)*(121)} = \sqrt{-1} * \sqrt{2}*\sqrt{121} =\pm11i\sqrt2$

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Example Question #1972 : Mathematical Relationships And Basic Graphs

Simplify:

(1+2i)(4-3i)

Possible Answers:

-2+5i

4-i

10-5i

-2-5i

10+5i

Correct answer:

10+5i

Explanation:

To simplify, we must FOIL:

(1+2i)(4-3i)=4-3i+8i-6i^2

Remembering that $i=\sqrt{-1}, i^2=-1$ , we can simplify this further to

4+5i-6(-1)=10+5i

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Example Question #23 : Complex Imaginary Numbers

Simplify: $i^{348}+i^{-348}$

Possible Answers:

$\frac{1}{i}$

Correct answer:

Explanation:

Rewrite the expression and eliminate the negative exponent.

$i^{348}+\frac{1}{i^{348}}$

Evaluate $i^{348}$ .

$i=\sqrt{-1}$

$i^2 = \sqrt{-1}\cdot \sqrt{-1} = -1$

$i^3 = i\cdot i^2 = -i$

$i^4 = i^2\cdot i^2 = -1\cdot -1 =1$

We can rewrite $i^{348}$ as an exponential product using i^4 .

$i^{348} = (i^4)^{87} = (1)^{87} = 1$

Rewrite the expression.

$i^{348}+\frac{1}{i^{348}} = 1+\frac{1}{1} = 2$

The answer is:

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Example Question #24 : Complex Imaginary Numbers

$i^{1680}$

Possible Answers:

i+1

Correct answer:

Explanation:

i raised to each successive power has a different answer that repeats in a cycle of four.

$i^{1}=i$

$i^{2}=-1$

$i^{3}=-i$

$i^{4}=1$

Then, the cycle repeats with $i^{5}=i$ and $i^{6}=-1$ and so on.

To solve any problem with i raised to a large number, the easiest way is to find the closest number that either is the exponent or is under the exponent that evenly divides by four and completes the cycle.

In this problem

$i^{1680}$

The number 1680 divides evenly by 4. This means that 1680 completes the cycle and has the same answer as $i^{4}$

$i^{1680}=1$

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Algebra II : Mathematical Relationships and Basic Graphs

Study concepts, example questions & explanations for Algebra II

All Algebra II Resources

Example Questions

Example Question #11 : Sat Subject Test In Math I

Example Question #31 : Imaginary Numbers & Complex Functions

Example Question #1 : Complex Conjugates

Example Question #15 : Number Theory

Example Question #12 : Imaginary Numbers

Example Question #13 : Imaginary Numbers

Example Question #1971 : Mathematical Relationships And Basic Graphs

Example Question #1972 : Mathematical Relationships And Basic Graphs

Example Question #23 : Complex Imaginary Numbers

Example Question #24 : Complex Imaginary Numbers

All Algebra II Resources

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