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Algebra II : Algebra II

Study concepts, example questions & explanations for Algebra II

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10 Diagnostic Tests 630 Practice Tests Question of the Day Flashcards Learn by Concept

Example Questions

Example Question #3 : Imaginary Roots Of Negative Numbers

Simplify:

(2+5i)(-8-i)

Possible Answers:

-18-5i

12+6i

42+11i

-11-42i

Correct answer:

-11-42i

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)(-8-i)=-16-2i-40i-5i^2

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

Substitute in for i^2

-16-2i-40i-5i^2=-16-42i-5(-1)=-16+5-42i=-11-42i

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Example Question #4 : Imaginary Roots Of Negative Numbers

Simplify:

(2+6i)(5+3i)

Possible Answers:

12-8i

-36-8i

-8+36i

36-8i

Correct answer:

-8+36i

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+6i)(5+3i)=10+6i+30i+18i^2

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

Substitute in for i^2

10+6i+30i+18i^2=10+36i+18(-1)=10-18+36i=-8+36i

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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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Algebra II : Algebra II

Study concepts, example questions & explanations for Algebra II

All Algebra II Resources

Example Questions

Example Question #3 : Imaginary Roots Of Negative Numbers

Example Question #4 : Imaginary Roots Of Negative Numbers

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

All Algebra II Resources

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