Imaginary Roots of Negative Numbers - GRE Subject Test: Math

Card 0 of 14

Question

Evaluate:

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

Answer

We can set in the cube of a binomial pattern:

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

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

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

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

Compare your answer with the correct one above

Question

Solve for and :

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

Answer

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:

No matter how you solve it, you get the values , .

Compare your answer with the correct one above

Question

Simplify the following product:

Answer

Multiply these complex numbers out in the typical way:

${}(5+3i)(-2+i) = -10+5i-6i+3i^2$

and recall that by definition. Then, grouping like terms we get

which is our final answer.

Compare your answer with the correct one above

Question

Simplify:

Answer

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.

Remember that $i=\sqrt-1$ , so .

Substitute in for

Compare your answer with the correct one above

Question

Simplify:

Answer

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.

Remember that $i=\sqrt-1$ , so .

Substitute in for

Compare your answer with the correct one above

Question

Simplify:

Answer

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.

Remember that $i=\sqrt-1$ , so .

Substitute in for

Compare your answer with the correct one above

Question

Simplify: $\sqrt{-243}$

Answer

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

We need to multiply everything that we bring outside:

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: $9i\sqrt{3}$

Compare your answer with the correct one above

Question

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

Answer

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:

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

It checks out, so the complex root is

Compare your answer with the correct one above

Question

$Simplify: \sqrt{-72}$

Answer

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

$i=\sqrt{-1}$

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

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

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

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

Compare your answer with the correct one above

Question

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

Answer

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:

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

$\sqrt{i^2}=i$

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

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

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

Compare your answer with the correct one above

Question

$Simplify:\ \sqrt{-150}$

Answer

$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.

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

Compare your answer with the correct one above

Question

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

Answer

$First\ simplify\ each\ radical:$

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

$Next\ combine\ like\ terms:$

Compare your answer with the correct one above

Question

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

Answer

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

We will group the 4 terms into two binomials:

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

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

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

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

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

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

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

Compare your answer with the correct one above

Question

$Simplify: \sqrt{-98}$

Answer

$Simplify: \sqrt{-98}$

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

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

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

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

Compare your answer with the correct one above

Tap the card to reveal the answer