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SAT II Math I : Irrational Numbers

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 #1 : Irrational Numbers

Which of the following is not an irrational number?

Possible Answers:

$\sqrt[5]{125}$

$\sqrt[3]{125}$

$\sqrt[4]{125}$

$\sqrt[6]{125}$

$\sqrt[2]{125}$

Correct answer:

$\sqrt[3]{125}$

Explanation:

A root of an integer is one of two things, an integer or an irrational number. By testing all five on a calculator, only $\sqrt[3]{125}$ comes up an exact integer - 5. This is the correct choice.

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Example Question #2 : Irrational Numbers

Simplify by rationalizing the denominator:

$\frac{45}{5 + \sqrt{7}}$

Possible Answers:

$\frac{ 25 + 5 \sqrt{7} }{ 2}$

$\frac{ 25 - \sqrt{7} }{ 2}$

$\frac{ 25 - 5 \sqrt{7} }{ 2}$

$\frac{15 \sqrt{7}}{4}$

$\frac{ 25 + \sqrt{7} }{ 2}$

Correct answer:

$\frac{ 25 - 5 \sqrt{7} }{ 2}$

Explanation:

Multiply the numerator and the denominator by the conjugate of the denominator, which is $5 - \sqrt{7}$ . Then take advantage of the distributive properties and the difference of squares pattern:

$\frac{45}{5 + \sqrt{7}}$

$= \frac{45\left (5 - \sqrt{7} \right )}{\left (5 + \sqrt{7} \right )\left (5 - \sqrt{7} \right )}$

$= \frac{45\left (5 - \sqrt{7} \right )}{ 5 ^{2}- \left ( \sqrt{7} \right ) ^{2} }$

$= \frac{45\left (5 - \sqrt{7} \right )}{ 25 -7}$

$= \frac{45\left (5 - \sqrt{7} \right )}{ 18}$

$= \frac{5\left (5 - \sqrt{7} \right )}{ 2}$

$= \frac{ 25 - 5 \sqrt{7} }{ 2}$

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Example Question #1 : Irrational Numbers

$\frac{-5+10i}{3+4i}=?$

Possible Answers:

1+2i

-1+2i

$-\frac{5}{3}+\frac{5}{2}i$

1-2i

$-\frac{5}{3}-\frac{5}{2}i$

Correct answer:

1+2i

Explanation:

$\frac{-5+10i}{3+4i}$

$=\frac{(-5+10i)(3-4i)}{(3+4i)(3-4i)}$

$=\frac{-15+20i+30i-40i^{2}}{9-16i^{2}}$

$=\frac{-15+50i+40}{9+16}$

$=\frac{25+50i}{25}$

=1+2i

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Example Question #1 : Imaginary Numbers

Multiply:

(7 + 3i) (1 - 2i)

Possible Answers:

13 - 11i

7 - 5i

1 - 11i

13 - 17i

7 - 17i

Correct answer:

13 - 11i

Explanation:

Use the FOIL technique:

(7 + 3i) (1 - 2i)

$= 7 \cdot 1 - 7 \cdot 2i + 3i \cdot 1 - 3i \cdot 2i$

$= 7 - 14i + 3i - 6i ^{2}$

= 7 - 14i + 3i - 6(-1)

= 7 - 14i + 3i +6

= 13 - 11i

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Example Question #61 : Complex Imaginary Numbers

Evaluate:

$\left (2 + 3i \right )^{3}$

Possible Answers:

-46 + 63 i

-62 + 63 i

-4 6+ 9i

8 + 9i

8 - 27i

Correct answer:

-4 6+ 9i

Explanation:

We can set A = 2, B = 3i in the cube of a binomial pattern:

$\left ( A + B\right ) ^{3} = A ^{3} + 3 A^{2 }B + 3AB^{2} + B^{3}$

$= 2 ^{3} + 3 \cdot 2 ^{2 } \cdot 3i + 3 \cdot 2 \cdot (3i)^{2} + (3i)^{3}$

$= 2^{3} + 3 \cdot 2^{2} \cdot 3i + 3 \cdot 2 \cdot 3 ^{2}\cdot i^{2} + 3^{3} \cdot i^{3}$

$= 8 + 36i + 54 i^{2} + 27 i^{3}$

= 8 + 36i + 54 (-1) + 27 (-i)

= 8 + 36i - 54 - 27 i

= -46 + 9i

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Example Question #1 : Imaginary Numbers

Evaluate $\small \frac{-6+2i}{10-3i}$

Possible Answers:

You cannot divide by complex numbers

$\small \small \frac{-66+2i}{91-60i}$

$\small \frac{-54+38i}{109}$

$\small \small \frac{-54+38i}{91-60i}$

$\small \frac{-66+2i}{109}$

Correct answer:

$\small \frac{-66+2i}{109}$

Explanation:

To divide by a complex number, we must transform the expression by multiplying it by the complex conjugate of the denominator over itself. In the problem, $\small 10-3i$ is our denominator, so we will multiply the expression by $\small \frac{10+3i}{10+3i}$ to obtain:

$\small \frac{-6+2i}{10-3i}\frac{10+3i}{10+3i}=\frac{-60+20i-18i+6i^2}{100-30i+30i-9i^2}$ .

We can then combine like terms and rewrite all $\small i^2$ terms as $\small -1$ . Therefore, the expression becomes:

$\small \frac{-60+2i+6i^2}{100-9i^2}=\frac{-66+2i}{109}$

Our final answer is therefore $\small \frac{-66+2i}{109}$

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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 #1 : Complex 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 #1 : Complex Imaginary Numbers

Simplify:

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

Possible Answers:

-3+4i

3-4i

11-4i

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 #3 : Irrational Numbers

Simplify:

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

Possible Answers:

7+16i

16+32i

25+8i

46+38i

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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SAT II Math I : Irrational Numbers

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

All SAT II Math I Resources

Example Questions

Example Question #1 : Irrational Numbers

Example Question #2 : Irrational Numbers

Example Question #1 : Irrational Numbers

Example Question #1 : Imaginary Numbers

Example Question #61 : Complex Imaginary Numbers

Example Question #1 : Imaginary Numbers

Example Question #4621 : Algebra Ii

Example Question #1 : Complex Imaginary Numbers

Example Question #1 : Complex Imaginary Numbers

Example Question #3 : Irrational Numbers

All SAT II Math I Resources

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