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GRE Subject Test: Math : Algebra

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

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2 Diagnostic Tests 148 Practice Tests Question of the Day Flashcards Learn by Concept

Example Questions

Example Question #81 : Classifying Algebraic Functions

Simplify: (13-7i)-(10-8i)

Possible Answers:

3-15i

-74-174i

23-i

3+i

Correct answer:

3+i

Explanation:

$When\ subtracting\ binomials\ in\ ( )\; first\ distribute\ the\ negative\ sign.$

(13-7i)-(10-8i)=13-7i-10+8i

$Now\ combine\ like\ terms.$

13-10-7i+8i=3+i

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Example Question #4 : Operations On Complex Numbers

What is the value of (3-4i)(2i+3) ?

Possible Answers:

None of the other answers

-6i+17

-6i+1

-6i-17

6i-17

Correct answer:

-6i+17

Explanation:

Distribute and Multiply:

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

Simplify all terms...

6i+9-8(-1)-12i

=6i+9+8-12i

=-6i+17

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Example Question #5 : Operations On Complex Numbers

What is the value: $i^{24}(i^{3}+i^{-2})$ ?

Possible Answers:

-i-1

1+i

-1-i

-(-i+1)

Correct answer:

-1-i

Explanation:

Step 1: Recall the cycle of imaginary numbers to a random power .

If x=1 , then i^1=i

If x=2 , then i^2=-1

If x=3 , then i^3=-i

If x=4 , then i^4=1

If x=5 , then i^5=i^4(i^1)=1(i)=i

and so on....

The cycle repeats every terms.

For ANY number $x\ge 5$ , you can break down that term into smaller elementary powers of i.

Step 2: Distribute the $i^{24}$ to all terms in the parentheses:

$(i^{24})(i^3)+(i^{24})(i^{-2})$ .

Step 3: Recall the rules for exponents:

$\forall\{a,b,c,d\} \in \mathbb{R}::$

$(a^b)\cdot (a^c)=a^{b+c}$

${a^{-b}}=\frac {1}{a^b}$

$(a^b)\cdot (a^{-c})=a^b \cdot \bigg (\frac {1}{a^c}\bigg)=\frac {a^b}{a^c}=a^{b-c}=a^d$

Step 4: Use the rules to rewrite the expression in Step 2:

$(i^{24})\cdot {i^3}=i^{24+3}=i^{27}$

$(i^{24})(i^{-2})=\frac {i^{24}}{i^2}=i^{24-2}=i^{22}$

Step 5: Simplify the results in Step 4. Use the rules in Step 1.:

$i^{27}=(i^{24}\cdot i^3)=1(-i)=-i$

$i^{22}=(i^{20}\cdot i^{2})=1(-1)=-1$

Step 6: Write the answer in a+bi form, where is the real part and is the imaginary part:

We get -1-i

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Example Question #82 : Classifying Algebraic Functions

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

Possible Answers:

1-12i

5+2i

6-35i

1+12i

Correct answer:

5+2i

Explanation:

When adding complex numbers, we add the real numbers and add the imaginary numbers.

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

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

=5+2i

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Example Question #83 : Classifying Algebraic Functions

Subtract: (15+7i)-(12+9i)

Possible Answers:

3-2i

3-16i

3-5i

3+16i

Correct answer:

3-2i

Explanation:

In order to subtract complex numbers, we must first distribute the negative sign to the second complex number.

Subtract: (15+7i)-(12+9i)

15+7i-12-9i

15-12+7i-9i=3-2i

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Example Question #1 : Operations On Complex Numbers

Simplify: 3i(-5+2i)

Possible Answers:

15+6i^2

15-6i

6i^2-15i

-6-15i

Correct answer:

-6-15i

Explanation:

Simplify: 3i(-5+2i)

First we must distribute

3i(-5+2i)=-15i+2i^2

$We\ must\ remember\ that\ i^2=-1$

-15i+6i^2=-15i+6(-1)

$-15i+6(-1)=-6-15i\ (i\ term\ is\ always\ last)$

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Example Question #11 : Operations On Complex Numbers

Simplify: (4-7i)(1-2i)

Possible Answers:

-10-15i

5-9i

4-15i+14i^2

4-14i

Correct answer:

-10-15i

Explanation:

$In\ order\ to\ simplify\ (4-7i)(1-2i)\ we\ must\ multiply\ the\ binomials\ using\ the\ FOIL\ method.$

$F\ is\ for\ first\ times\ first: =4*1=4$

$O\ is\ for\ outside\ times\ outside =4*(-2i)=-8i$

$I\ is\ for\ inside\ times\ inside=(-7i)*1=-7i$

$L\ is\ for\ last\ times\ last =(-7i)*(-2i)=14i^2$

$Now\ we\ combine\ like\ terms\ and\ simplify:$

4-8i-7i+14i^2

$^* Remember\ that\ i^2=-1$

4-15i+14(-1)=-10-15i

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Example Question #84 : Classifying Algebraic Functions

Simplify: (5-8i)^2

Possible Answers:

25-64i

25-64i^2

-39-80i

Correct answer:

-39-80i

Explanation:

$When\ a\ binomial\ is\ squared\ we\ must\ write\ it\ out\ twice\ and\ perform\ the\ FOIL\ method.$

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

$F\ is\ for\ first\ times\ first = 5*5=25$

$O\ is\ for\ outside\ times\ outside = 5*8i=40i$

$I\ is\ for\ inside\ times\ inside: 8i*5=40i$

$L\ is\ for\ last\ times\ last = 8i*8i=64i^2=64(-1)=-64$

Now we put each of these together and combine like terms:

25-40i-40i-64=-39-80i

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Example Question #21 : Imaginary Numbers & Complex Functions

$\frac{\sqrt{-16}}{\sqrt{-8}}$

Possible Answers:

$-\sqrt{2}$

$\sqrt{2}$

Correct answer:

$\sqrt{2}$

Explanation:

$\frac{\sqrt{-16}}{\sqrt{-8}}$

Take i (the square root of -1) out of both radicals then divide.

$\frac{i\sqrt{16}}{i\sqrt{8}}$

$\frac{\sqrt{16}}{\sqrt{8}}$

$\sqrt{2}$

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Example Question #272 : Gre Subject Test: Math

$\sqrt{-16}\cdot \sqrt{-25}$

Possible Answers:

40i

$4\sqrt{5}$

20i

Correct answer:

Explanation:

$\sqrt{-16}\cdot \sqrt{-25}$

Take out i (the square root of -1) from both radicals and then multiply. You are not allowed to first multiply the radicals and then simplify because the roots are negative.

$4i\cdot 5i$

$20i^{2}$

Make i squared -1

20(-1)

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GRE Subject Test: Math : Algebra

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

All GRE Subject Test: Math Resources

Example Questions

Example Question #81 : Classifying Algebraic Functions

Example Question #4 : Operations On Complex Numbers

Example Question #5 : Operations On Complex Numbers

Example Question #82 : Classifying Algebraic Functions

Example Question #83 : Classifying Algebraic Functions

Example Question #1 : Operations On Complex Numbers

Example Question #11 : Operations On Complex Numbers

Example Question #84 : Classifying Algebraic Functions

Example Question #21 : Imaginary Numbers & Complex Functions

Example Question #272 : Gre Subject Test: Math

All GRE Subject Test: Math Resources

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