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Algebra II : Algebra II

Study concepts, example questions & explanations for Algebra II

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Example Questions

Example Question #2089 : Mathematical Relationships And Basic Graphs

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

Possible Answers:

$\frac{1}{3}$

i-2

2-i

$-\frac{1}{3}$

Correct answer:

$-\frac{1}{3}$

Explanation:

Simplify each term. Recall that $i=\sqrt{-1}$ .

$\frac{2}{i^4}+\frac{i^2}{3}-2 = \frac{2}{1}+\frac{-1}{3}-2$

Simplify the expression.

The answer is: $-\frac{1}{3}$

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Example Question #131 : Imaginary Numbers

Solve: -6i^2+30i^2-2i^2

Possible Answers:

Correct answer:

Explanation:

Rewrite the terms with the actual value of the imaginary term.

Recall that $i=\sqrt{-1}$ .

This means that: i^2 =-1

Replace this for all the terms.

-6i^2+30i^2-2i^2= -6(-1)+30(-1)-2(-1)

-6(-1)+30(-1)-2(-1) = -22

The answer is:

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Example Question #4751 : Algebra Ii

Evaluate: $-2i^{5000}$

Possible Answers:

-2-i

Correct answer:

Explanation:

Write the first few values of the imaginary term .

$i=\sqrt{-1}$

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

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

Rewrite the expression using the product of exponents.

$-2i^{5000} = -2(i^4)^{1250}= -2(1)^{1250}$

The value of one to this power will remain the same.

-2(1) =-2

The answer is:

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Example Question #2092 : Mathematical Relationships And Basic Graphs

Compute: 2i^2+3i^4-4i^6-5i^2

Possible Answers:

Correct answer:

Explanation:

To be able to evaluate the expression, we will need to write out the value of the imaginary terms.

Recall that: $i=\sqrt{-1}$

i^2= -1

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

$i^6 = i^4 \cdot i^2 = (1) \cdot (-1) = -1$

Replace the terms.

2i^2+3i^4-4i^6-5i^2= 2(-1)+3(1)-4(-1)-5(-1)

=-2+3+4+5 = 10

The answer is:

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Example Question #4751 : Algebra Ii

Evaluate: (3+2i)^3

Possible Answers:

54-9i

9-46i

54i-9

46i-9

Correct answer:

46i-9

Explanation:

Rewrite the problem as separate groups of binomials.

(3+2i)^3=(3+2i)(3+2i)(3+2i)

Use the FOIL method to expand the first two terms.

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

Simplify the right side.

9+12i+4i^2

Recall that since $i=\sqrt{-1}$ , the value of $i^2 = i\cdot i =-1$ .

9+12i+4(-1) = 12i+5

Multiply this value with the third binomial.

(12i+5)(3+2i) = (12i)(3)+(12i)(2i)+(5)(3)+(5)(2i)

Simplify the terms.

36i+24i^2+15+10i=36i+24(-1)+15+10i = 46i-9

The answer is: 46i-9

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Example Question #4752 : Algebra Ii

Solve: $\frac{-2i^3}{i^{-7}}-i^2$

Possible Answers:

$\textup{The answer is not given.}$

Correct answer:

Explanation:

In order to solve this expression, we will need to evaluate each term.

Use the property of exponents to subtract the powers in the fraction.

$\frac{-2i^3}{i^{-7}} = -2i^{3-(-7)} = -2i^{10}$

This term can be written as a fraction.

$-2i^{-10} = -2\cdot \frac{1}{i^{10}}$

Recall that the imaginary term $i=\sqrt{-1}$ . This means that:

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

$i^{10} = (i^2)^5 = (-1)^5 = -1$

Replace the terms.

$-2\cdot \frac{1}{i^{10}} = -2\cdot \frac{1}{-1} = 2$

Solve the expression by replacing the values in the original expression.

$\frac{-2i^3}{i^{-7}}-i^2 = 2-(-1) = 2+1 =3$

The answer is:

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Example Question #4753 : Algebra Ii

Simplify: $-1000i^{52}+ 500i^{30}$

Possible Answers:

1000-500i

1000+500i

-500i

Correct answer:

Explanation:

Write the first few powers of the imaginary term.

$i=\sqrt{-1}$

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

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

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

We can then rewrite the higher powered imaginary terms by the product of exponents.

$-1000i^{52}+ 500i^{30} = -1000(i^4)^{13}+500(i^2)^{15}$

Simplify the terms.

$-1000(1)^{13}+500(-1)^{15} = -1000-500 = -1500$

The answer is:

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Example Question #2096 : Mathematical Relationships And Basic Graphs

Compute: $-5i^{1234}$

Possible Answers:

-5-i

-5i

Correct answer:

Explanation:

Identify the first two powers of the imaginary term.

$i=\sqrt{-1}$

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

Rewrite the expression as a product of exponents.

$-5i^{1234}=-5(i^{2})^{617} = -5(-1)^{617}$

Negative one to an odd power will be negative one.

-5(-1) =5

The answer is:

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Example Question #2097 : Mathematical Relationships And Basic Graphs

Evaluate: $2i^{878}$

Possible Answers:

$\textup{The answer is not given.}$

-2i

Correct answer:

Explanation:

Write the first few powers of the imaginary term.

$i=\sqrt{-1}$

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

Change the higher ordered power by using the power rule of exponents.

$2i^{878} = 2(i^2)^{439}=2(-1)^{439}=2(-1)$

A negative one to an odd power will be negative one.

The answer is:

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Example Question #2098 : Mathematical Relationships And Basic Graphs

Add $8 +2 i \sqrt{5}$ to its complex conjugate. What is the result?

Possible Answers:

$-4i \sqrt{5}$

$4i \sqrt{5}$

Correct answer:

Explanation:

The complex conjugate of a complex number a+bi is a-bi .

Therefore, the complex conjugate of $8 +2 i \sqrt{5}$ is $8 -2 i \sqrt{5}$ . Add the two:

$\left (8 +2 i \sqrt{5} \right )+\left ( 8 -2 i \sqrt{5} \right )$

Collect real parts and imaginary parts:

$=8+ 8 +2 i \sqrt{5} -2 i \sqrt{5}$

The imaginary parts cancel out:

=8+ 8

=16

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Algebra II : Algebra II

Study concepts, example questions & explanations for Algebra II

All Algebra II Resources

Example Questions

Example Question #2089 : Mathematical Relationships And Basic Graphs

Example Question #131 : Imaginary Numbers

Example Question #4751 : Algebra Ii

Example Question #2092 : Mathematical Relationships And Basic Graphs

Example Question #4751 : Algebra Ii

Example Question #4752 : Algebra Ii

Example Question #4753 : Algebra Ii

Example Question #2096 : Mathematical Relationships And Basic Graphs

Example Question #2097 : Mathematical Relationships And Basic Graphs

Example Question #2098 : Mathematical Relationships And Basic Graphs

All Algebra II Resources

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