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GRE Subject Test: Math : Imaginary Roots of Negative Numbers

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Example Questions

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Example Question #71 : Classifying Algebraic Functions

$Simplify:\ \sqrt{-150}$

Possible Answers:

$-5\sqrt{6}$

$5i\sqrt{3}$

$-5\sqrt{3}$

$5i\sqrt{6}$

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 #251 : Gre Subject Test: Math

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

Possible Answers:

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

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

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

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

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 #252 : Gre Subject Test: Math

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

Possible Answers:

-15i

225i

-225i

15i

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 #11 : Imaginary Numbers & Complex Functions

$Simplify: \sqrt{-98}$

Possible Answers:

$7\sqrt{2i}$

$-7\sqrt{2}$

$7\sqrt{-2}$

$7i\sqrt{2}$

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 : Imaginary Roots of Negative Numbers

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

All GRE Subject Test: Math Resources

Example Questions

Example Question #71 : Classifying Algebraic Functions

Example Question #251 : Gre Subject Test: Math

Example Question #252 : Gre Subject Test: Math

Example Question #11 : Imaginary Numbers & Complex Functions

All GRE Subject Test: Math Resources

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