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GRE Subject Test: Math : GRE Subject Test: Math

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Example Questions

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 #1 : Equations With Complex Numbers

Solve for and :

$ai^{93}+bi^{35}+ai^{24}-bi^{86}=40+20i$

Possible Answers:

a=30, b=10

a=15, b=15

a=40, b=20

a=20, b=40

a=10, b=30

Correct answer:

a=30, b=10

Explanation:

Remember that

$i = \sqrt{-1}\\ i^2 = \sqrt{-1}^2 = -1\\ i^3 = i^2\cdot i = -1 \cdot i = -i\\ i^4 = i^2 \cdot i^2 = (-1)\cdot(-1) = 1\\ i^5 = i^4\cdot i = 1\cdot i = i = \sqrt{-1}\\ i^6 = i^4\cdot i^2 = 1\cdot i^2 = i^2 =-1$

So the powers of are cyclic. This means that when we try to figure out the value of an exponent of , we can ignore all the powers that are multiples of because they end up multiplying the end result by , and therefore do nothing.

This means that

$ai^{93} + bi^{35} + ai^{24} - bi^{86}\\ = ai^{92+1} + bi^{32+3} + ai^{24+0} - bi^{84+2}\\ = ai^{92}\cdot i^1 + bi^{32}\cdot i^3 + ai^{24}\cdot i^0 - bi^{84}\cdot i^2\\ =a(i^4)^{23}\cdot i^1 + b(i^4)^8\cdot i^3 + a(i^4)^6\cdot i^0 - b(i^4)^{21}\cdot i^2\\ = a (1^{23})\cdot i + b(1^8)\cdot i^3 + a(1^6) - b(1^{21})\cdot i^2$

Now, remembering the relationships of the exponents of , we can simplify this to:

$ai - bi + a - (-b) = ai-bi+a+b\\ = (a+b) + (a-b)i = 40+20i$

Because the elements on the left and right have to correspond (no mixing and matching!), we get the relationships:

a+b=40

a-b=10

No matter how you solve it, you get the values a=30 , b=10 .

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Example Question #1 : Imaginary Roots Of Negative Numbers

Simplify: $\sqrt{-243}$

Possible Answers:

None of the Above

$-9i\sqrt{3}$

$9i\sqrt{3}$

Correct answer:

$9i\sqrt{3}$

Explanation:

Step 1: Split the $\sqrt{-243}$ into $\sqrt{243}\cdot\sqrt{-1}$ .

Step 2: Recall that $\sqrt{-1}=i$ , so let's replace it.

We now have: $\sqrt{243}\cdot i$ .

Step 3: Simplify $\sqrt{243}$ . To do this, we look at the number on the inside.

$243=3\cdot3\cdot3\cdot3\cdot3$ .

Step 4: Take the factorization of 243 and take out any pairs of numbers. For any pair of numbers that we find, we only take of the numbers out.

We have a pair of 3's , so a is outside the radical.
We have another pair of 3's , so one more three is put outside the radical.

We need to multiply everything that we bring outside: 3*3=9

Step 5: The goes with the 9...

Step 6: The last after taking out pairs gets put back into a square root and is written right after the

It will look something like this: $9i\sqrt{3}$

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Example Question #3 : Imaginary Numbers & Complex Functions

$Simplify: 5\sqrt{12i^{2}}$

Possible Answers:

$5i\sqrt{12}$

$2i\sqrt{3}$

$5\sqrt{-12}$

$10i\sqrt{}3$

Correct answer:

$10i\sqrt{}3$

Explanation:

There are two ways to simplify this problem:

Method 1:

$We\ know\ that\ i^2=-1,\ when\ we\ replace\ that\ we\ get:$

$5\sqrt{-12}$

$=5i\sqrt{12}(take\ the\ i\ out)$

$=5i\sqrt{4}\sqrt{3} (the\ perfect\ square\ 4\ will\ go\ into\ 12)$

$=10i{\sqrt{3}}\ reduce$

Method 2:

$We\ know\ that\ the\ square\ root\ of\ anything\ squared\ is\ just\ itself$

$\sqrt{i^2}=i$

$5\sqrt{12i^{2}}=5i\sqrt{12}$

$=5i\sqrt{4}\sqrt{3} (the\ perfect\ square\ 4\ will\ go\ into\ 12)$

$=10i{\sqrt{3}}\ reduce$

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Example Question #71 : Algebra

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

Possible Answers:

4-7i

7-4i

-11i

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 #11 : Imaginary Roots Of Negative Numbers

$Simplify:\ \sqrt{-150}$

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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GRE Subject Test: Math : GRE Subject Test: Math

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

All GRE Subject Test: Math Resources

Example Questions

Example Question #4 : Imaginary Roots Of Negative Numbers

Example Question #1 : Equations With Complex Numbers

Example Question #1 : Imaginary Roots Of Negative Numbers

Example Question #3 : Imaginary Numbers & Complex Functions

Example Question #71 : Algebra

Example Question #6 : Imaginary Roots Of Negative Numbers

Example Question #11 : Imaginary Roots Of Negative Numbers

Example Question #12 : Imaginary Roots Of Negative Numbers

Example Question #11 : Imaginary Numbers & Complex Functions

Example Question #12 : Imaginary Numbers & Complex Functions

All GRE Subject Test: Math Resources

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