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SAT II Math I : Number Theory

Study concepts, example questions & explanations for SAT II Math I

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All SAT II Math I Resources

6 Diagnostic Tests 113 Practice Tests Question of the Day Flashcards Learn by Concept

Example Questions

Example Question #4621 : Algebra Ii

Simplify the following product:

{} (5+3i)(-2+i)

Possible Answers:

{}-7-i

{} -13-i

{} -10+3i

{} -13+11i

Correct answer:

{} -13-i

Explanation:

Multiply these complex numbers out in the typical way:

${}(5+3i)(-2+i) = -10+5i-6i+3i^2$

and recall that i^2=-1 by definition. Then, grouping like terms we get

{} (-10-3)+(5i-6i) = -13-i

which is our final answer.

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Example Question #2 : Imaginary Numbers

Identify the real part of 1-3i

Possible Answers:

none of the above.

Correct answer:

Explanation:

A complex number in its standard form is of the form: a+bi , where stands for the real part and stands for the imaginary part. The symbol stands for $\sqrt{-1}$ .

The real part in this problem is 1.

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Example Question #11 : Sat Subject Test In Math I

Simplify:

(4-6i)+(7+2i)

Possible Answers:

3-4i

11-4i

-3+4i

5+2i

Correct answer:

11-4i

Explanation:

To add complex numbers, find the sum of the real terms, then find the sum of the imaginary terms.

(4-6i)+(7+2i)=(4+7)+(-6+2)i=11-4i

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Example Question #2 : Imaginary Roots Of Negative Numbers

Simplify:

(5+8i)(6-2i)

Possible Answers:

16+32i

46+38i

25+8i

7+16i

Correct answer:

46+38i

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.

(5+8i)(6-2i)=30-10i+48i-16i^2

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

Substitute in for i^2

30-10i+48i-16i^2=30+38i-16(-1)=30+38i+16=46+38i

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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 #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 #2 : Understanding Imaginary And Complex Numbers

Simplify the expression.

4i^2-6i-7i^2+3i+4

Possible Answers:

-3i^2-3i+4

None of the other answer choices are correct.

-7-3i

7-3i

Correct answer:

7-3i

Explanation:

Combine like terms. Treat $\small i$ as if it were any other variable.

4i^2-6i-7i^2+3i+4

-3i^2-3i+4

Substitute to eliminate $\small i^2$ .

i^2=-1

-3(-1)-3i+4

Simplify.

3-3i+4=7-3i

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SAT II Math I : Number Theory

Study concepts, example questions & explanations for SAT II Math I

All SAT II Math I Resources

Example Questions

Example Question #4621 : Algebra Ii

Example Question #2 : Imaginary Numbers

Example Question #11 : Sat Subject Test In Math I

Example Question #2 : Imaginary Roots Of Negative Numbers

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 #1 : Complex Conjugates

Example Question #15 : Number Theory

Example Question #2 : Understanding Imaginary And Complex Numbers

All SAT II Math I Resources

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