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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 #21 : Imaginary Numbers

Simplify $i^{358}$

Possible Answers:

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

$i^{358}$

The closest number under 358 that evenly divides by 4 is 356. 356 completes the cycle. 358 is two numbers into the start of a new cycle and has the same value as $i^{2}$

$i^{358}=-1$

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Example Question #25 : Complex Imaginary Numbers

Simplify $i^{273}$

Possible Answers:

$-i^{2}$

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.

In this problem

$i^{273}$

The closest number under 273 that divides by 4 is 272. This means the next number 273 will be the start of a new cycle and have the same answer as $i^{1}$

$i^{273}=i$

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Example Question #4641 : Algebra Ii

Simplify $i^{703}$

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.

In this problem

$i^{703}$

The closest number under the exponent that divides evenly by 4 is 700. This means that 703 is the third number in a new cycle and will have the same value as $i^{3}$

$i^{703}=-i$

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Example Question #27 : Complex Imaginary Numbers

Solve:

$((\sqrt{35})(i))^3$

Possible Answers:

$-35\sqrt{35}$

None of these

$-35i\sqrt{35}$

$-i\sqrt{35}$

$35i\sqrt{35}$

Correct answer:

$-35i\sqrt{35}$

Explanation:

Definition of

$i=\sqrt{-1}$

$i^3=\sqrt{-1}*\sqrt{-1}*\sqrt{-1}$

$i^3=-\sqrt{-1}$

i^3=-i

$(\sqrt{35})^3=35\sqrt{35}$

Thus,

$((\sqrt{35})(i))^3=-35i\sqrt{35}$

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Example Question #23 : Imaginary Numbers

Simplify the expression: $\frac{3-2i}{2-i}$

Possible Answers:

$\frac{8+i}{5}$

$\textup{The expression is simplified as is.}$

$\frac{8-i}{5}$

$\frac{i-8}{5}$

Correct answer:

$\frac{8-i}{5}$

Explanation:

Multiply the numerator and denominator by the conjugate of the denominator.

$\frac{3-2i}{2-i} \cdot \frac{2+i}{2+i}$

Multiply both the top and bottom by using the FOIL method.

Numerator:

(3-2i)(2+i) = (3)(2)+(3)(i)+(-2i)(2)+(-2i)(i)

=6+3i-4i-2i^2

=6-i-2i^2

Recall that $i=\sqrt{-1}$ , which means that i^2 = -1 .

=6-i-2(-1)= 6-i+2 = 8-i

Denominator:

(2-i)(2+i) = 4-i^2 = 4-(-1) = 5

Divide the numerator with the denominator.

The answer is: $\frac{8-i}{5}$

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Example Question #29 : Complex Imaginary Numbers

Simplify: $\frac{i-1}{i-2}$

Possible Answers:

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

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

$\frac{3+i}{5}$

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

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

Correct answer:

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

Explanation:

Multiply the top and bottom by the conjugate of the denominator.

$\frac{i-1}{i-2}\cdot \frac{i+2}{i+2}$

Multiply the numerator with the numerator. Use the FOIL method.

(i-1)(i+2) = i^2+(2)(i)-i-2 = i^2+i-2

Recall that $i=\sqrt{-1}$ , which indicates that i^2 = -1 .

i^2+i-2 =-1+i-2 = i-3

Multiply the denominator with the denominator.

(i-2)(i+2) = i^2-4 = -1-4 = -5

Divide the numerator with the denominator.

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

The answer is: $\frac{3-i}{5}$

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Example Question #31 : Imaginary Numbers

Solve: $\frac{6i}{2i-1}$

Possible Answers:

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

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

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

$\textup{The expression is fully simplified.}$

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

Correct answer:

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

Explanation:

In order to simplify this expression, we will need to multiply the numerator and denominator with the conjugate of the denominator.

$\frac{6i}{2i-1}\cdot \frac{2i+1}{2i+1}$

Simplify the top and bottom. The value of $i=\sqrt{-1}$ .

6i(2i+1) = 12i^2+6i = 12(-1)+6i = 6i-12

(2i-1)(2i+1) = 4i^2-1 = 4(-1)-1=-5

Divide the numerator with the denominator.

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

The answer is: $-\frac{6}{5}i+\frac{12}{5}$

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Example Question #32 : Imaginary Numbers

Simplify: $\frac{i^3}{2i-4}$

Possible Answers:

2i-4

$\frac{1}{5}i-\frac{1}{10}$

$\textup{The question is simplified as is.}$

$\frac{1}{2}i-\frac{1}{8}$

$i-\frac{1}{4}$

Correct answer:

$\frac{1}{5}i-\frac{1}{10}$

Explanation:

Multiply the numerator and denominator by the conjugate of the denominator.

$\frac{i^3}{2i-4}\cdot \frac{2i+4}{2i+4} = \frac{2i^4+4i^3}{4i^2-16}$

Simplify the top and the bottom.

Recall that $i=\sqrt{-1}$ , and $i^2= i\cdot i = -1$ .

This means that:

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

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

Re-substitute the actual values back into the fraction.

$\frac{2(1)+4(-i)}{4(-1)-16} = \frac{2-4i}{-20}$

Reduce and split this fraction.

The correct answer is: $\frac{1}{5}i-\frac{1}{10}$

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Example Question #31 : Imaginary Numbers

Simplify: $\frac{i^2+2i+1}{i-2}$

Possible Answers:

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

$\frac{1}{5}i-\frac{5}{6}$

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

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

$-\frac{4}{5}i+\frac{1}{5}$

Correct answer:

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

Explanation:

The values are imaginary. Recall that:

$i=\sqrt{-1}$

This means that: $i^2= i\cdot i = -1$

Rewrite the numerator.

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

Multiply the top and bottom by the conjugate of i-2 .

$\frac{2i}{i-2}\cdot \frac{i+2}{i+2} = \frac{2i^2+4i}{i^2-4}$

Rewrite the terms using the value of i^2 .

$\frac{2i^2+4i}{i^2-4}=\frac{2(-1)+4i}{-1-4} =\frac{ -2+4i}{-5}$

The answer is: $-\frac{4}{5}i+\frac{2}{5}$

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Example Question #34 : Imaginary Numbers

Simplify: $\frac{i-6}{2i-1}$

Possible Answers:

$\frac{11}{5}i+\frac{8}{5}$

11i-14

$\frac{1}{5}i-\frac{2}{5}$

-11i+14

$-\frac{13}{5}i-\frac{4}{5}$

Correct answer:

$\frac{11}{5}i+\frac{8}{5}$

Explanation:

In order to simplify this, we will need to multiply the top and bottom of the fraction by the conjugate of the denominator.

$\frac{i-6}{2i-1} \times \frac{2i+1}{2i+1}$

Multiply the numerator.

(i-6)(2i+1) = (i)(2i)+(i)(1)+(-6)(2i)+(-6)(1)

2i^2+i-12i-6 = 2i^2-11i-6

The term i^2=-1 since $i=\sqrt{-1}$ . Replace the term and simplify.

= 2(-1)-11i-6 = -11i-8

Multiply the denominator.

(2i-1)(2i+1) =4i^2-1 = 4(-1)-1 =-5

The expression becomes:

$\frac{i-6}{2i-1} \times \frac{2i+1}{2i+1} = \frac{ -11i-8}{-5}$

Dividing the double negatives will result into a positive value.

The answer is: $\frac{11}{5}i+\frac{8}{5}$

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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 #21 : Imaginary Numbers

Example Question #25 : Complex Imaginary Numbers

Example Question #4641 : Algebra Ii

Example Question #27 : Complex Imaginary Numbers

Example Question #23 : Imaginary Numbers

Example Question #29 : Complex Imaginary Numbers

Example Question #31 : Imaginary Numbers

Example Question #32 : Imaginary Numbers

Example Question #31 : Imaginary Numbers

Example Question #34 : Imaginary Numbers

All Algebra II Resources

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