Imaginary Roots of Negative Numbers

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GRE Quantitative Reasoning › Imaginary Roots of Negative Numbers

Questions 1 - 10
1

Explanation

2

Explanation

3

Explanation

There are two ways to simplify this problem:

Method 1:

Method 2:

4

Evaluate:

Explanation

We can set in the cube of a binomial pattern:

5

What are the imaginary root(s) of ?

Explanation

Rewrite the expression as a positive root and the negative root

Take the square root of the positive root:

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

6

Explanation

The perfect square of 25 will go into 150

The square root of 25 is 5.

7

Simplify:

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.

Remember that , so .

Substitute in for

8

Simplify:

None of the Above

Explanation

Step 1: Split the into .

Step 2: Recall that , so let's replace it.

We now have: .

Step 3: Simplify . To do this, we look at the number on the inside.

.

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:

9

Explanation

10

Explanation

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:

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