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GRE Subject Test: Math : Classifying Algebraic Functions

Study concepts, example questions & explanations for GRE Subject Test: Math

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Example Questions

Example Question #1 : Imaginary Numbers & Complex Functions

$Simplify: \sqrt{-49}-\sqrt{-16}$

Possible Answers:

7-4i

-11i

4-7i

Correct answer:

Explanation:

$First\ simplify\ each\ radical:$

$\sqrt{-49}-\sqrt{-16}=7i-4i$

$Next\ combine\ like\ terms:$

7i-4i=3i

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Example Question #6 : Imaginary Roots Of Negative Numbers

$Simplify: \sqrt{-72}$

Possible Answers:

$6{\sqrt{-2}}$

$-6i\sqrt{2}$

$6i{\sqrt{2}}$

$-6{\sqrt{2}}$

Correct answer:

$6i{\sqrt{2}}$

Explanation:

$When\ reducing\ the\ square\ root\ of\ a\ negative\ number\ -\ we\ must\ first\$

$i=\sqrt{-1}$

$\sqrt{-72}=\sqrt{-1}\sqrt{72}=i\sqrt{72}$

$Once\ we\ have\ taken\ out\ the\ i,\ then\ we\ can\ simply\ break\ down$ $the\ number\ under\ the\ radical\ by\ finding\ the\ largest\ perfect\ square\ that\ will go\ into\ it.$

$36\ will\ go\ into\ 72\ so\ we\ break\ this\ down\ to\ become:$

$i\sqrt{36}\sqrt{2}=6i\sqrt{2}$

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Example Question #71 : Algebra

$Simplify:\ \sqrt{-150}$

Possible Answers:

$5i\sqrt{3}$

$-5\sqrt{6}$

$-5\sqrt{3}$

$5i\sqrt{6}$

Correct answer:

$5i\sqrt{6}$

Explanation:

$When\ taking\ the\ square\ root\ of\ a\ negative\ number\ we\ must\ first\$

$i=\sqrt{-1}$

$\sqrt{-150}=\sqrt{-1}\sqrt{150}=i\sqrt{150}$

The perfect square of 25 will go into 150

$i\sqrt{150}=i\sqrt{25*6}$

The square root of 25 is 5.

$i\sqrt{25}\sqrt{6}=5i\sqrt{6}$

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Example Question #12 : Imaginary Roots Of Negative Numbers

$Find\ all\ roots\ for: f(x)=3x^3-x^2+15x-5$

Possible Answers:

$\frac{1}{3},\ and\ i\sqrt{5}$

$\1,\and\ i\sqrt{5}$

$\1,\ i\sqrt{5}\ and\ -i\sqrt{5}$

$\frac{1}{3},\ i\sqrt{5}\ and\ -i\sqrt{5}$

Correct answer:

$\frac{1}{3},\ i\sqrt{5}\ and\ -i\sqrt{5}$

Explanation:

In order to find all the roots for the polynomial, we must use factor by grouping:

f(x)=3x^3-x^2+15x-5

We will group the 4 terms into two binomials:

(3x^3-x^2)+(15x-5)

We then take the greatest common factor out of each binomial:

x^2(3x-1)+5(3x-1)

We can see now that each term has a common binomial as a factor:

(3x-1)(x^2+5)

In order to find the roots, we must set each factor equal to zero and solve:

$3x-1=0\ which\ gives\ 3x=1\ then\ x=\frac{1}{3}$

$x^2+5=0\ which\ gives\ x^2=-5\ which\ results\ in\ x=\sqrt{-5}$

$This\ will\ give\ us\ imaginary\ roots\ of\ x=\pm i\sqrt{5}$

$The\ three\ roots\ are\ \frac{1}{3},\ i\sqrt{5}\ and\ -i\sqrt{5}$

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Example Question #11 : Imaginary Numbers & Complex Functions

What are the imaginary root(s) of $\sqrt {-225}$ ?

Possible Answers:

-225i

-15i

15i

225i

Correct answer:

15i

Explanation:

Rewrite the expression as a positive root and the negative root

$\sqrt {-225}=\sqrt{225}\sqrt{-1}$

Take the square root of the positive root:

$\sqrt{225}\sqrt{-1}=15i$

To check the answer, square the square root:

(15i)^2=(15i)(15i)=(15^2)(i^2)=225(-1)=-225

should be what was inside the square root in the beginning.

It checks out, so the complex root is 15i

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Example Question #12 : Imaginary Numbers & Complex Functions

$Simplify: \sqrt{-98}$

Possible Answers:

$-7\sqrt{2}$

$7\sqrt{2i}$

$7i\sqrt{2}$

$7\sqrt{-2}$

Correct answer:

$7i\sqrt{2}$

Explanation:

$Simplify: \sqrt{-98}$

$In\ order\ to\ simplify\ the\ square\ root\ of\ a\ negative\ number\\ we\ must\ first\ take\ the\ i\ out.$

$We\ can\ replace\ the\sqrt{-98}\ with\sqrt{(-1)(2)(49)}$

$\sqrt{-1}=i\\ \ \ \ \sqrt{2}=does\ not\ reduce\\ \sqrt{49}=7$

$Combining\ all\ of\ the\ above\ will\ give\ us\ 7i\sqrt{2}$

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Example Question #1 : Operations On Complex Numbers

Expand and Simplify: (4-3i)(2+3i)(2-3i)

Possible Answers:

39+52i

-20-15i

52-39i

52+39i

Correct answer:

52-39i

Explanation:

Step 1: We will multiply the two complex conjugates: 2x-3 and 2x+3 .

(2-3i)(2+3i)=4+6i-6i-9i^2=4-9i^2

Step 2: Replace i^2 with .

4-9(-1)

Simplify:

4+9(1)
=13

Step 3: Multiply the result of the complex conjugates to the other parentheses, (4-3i) .

13(4-3i)=52-39i

The final answer after the product of all three binomials is 52-39i

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Example Question #1 : Operations On Complex Numbers

Expand: $(i-1)^{10}$ .

Possible Answers:

-32i

32i

32(1-i)

Correct answer:

-32i

Explanation:

Quick Way:
Step 1: Expand (i-1)^2 .

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

Remember: i^2=-1

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

Step 2: $(1-i)^{10}=(i-1)^{2\cdot 5}$

By this equivalence, I can just raise the answer of (1-i)^2 to the power .

$(i-1)^{10}=(-2i)^{5}=-32i^5$

$i^5 \equiv i^1=i$ . Replace i^5 ..

Final answer: -32i

Long Way:

Math work

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Example Question #1 : Operations On Complex Numbers

Multiply:

(2+3i)(3-4i)

Possible Answers:

6+i

18+i

18-i

-i-18

Correct answer:

18+i

Explanation:

Step 1: FOIL:

Recall, FOIL means to multiply the first terms in both binomials together, the outer terms together, the inner terms together, and finally, the last terms together.

(2+3i)(3-4i)=6-8i+9i-12i^2

Step 2: Simplify:

6+i-12i^2

Step 3: Recall: i^2=-1 . Replace and simplify.

6+i-12(-1)
6+i+12
=18+i

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Example Question #1 : Operations On Complex Numbers

Simplify: (3-5i)+(2+7i)

Possible Answers:

6-35i

41+11i

5+2i

Correct answer:

5+2i

Explanation:

When adding imaginary numbers, simply add the real parts and the imaginary parts.

(3-5i)+(2+7i)= (3+2)+(-5i+7i)=5+2i

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GRE Subject Test: Math : Classifying Algebraic Functions

Study concepts, example questions & explanations for GRE Subject Test: Math

All GRE Subject Test: Math Resources

Example Questions

Example Question #1 : Imaginary Numbers & Complex Functions

Example Question #6 : Imaginary Roots Of Negative Numbers

Example Question #71 : Algebra

Example Question #12 : Imaginary Roots Of Negative Numbers

Example Question #11 : Imaginary Numbers & Complex Functions

Example Question #12 : Imaginary Numbers & Complex Functions

Example Question #1 : Operations On Complex Numbers

Example Question #1 : Operations On Complex Numbers

Example Question #1 : Operations On Complex Numbers

Example Question #1 : Operations On Complex Numbers

All GRE Subject Test: Math Resources

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