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ACT Math 2.2 Intermediate Algebra
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ACT Math: Intermediate Algebra Drill 4, Problem 2. What is the root of this equation?

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SAT Math 6.2 Algebra and Functions 268 Views


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SAT Math 6.2 Algebra and Functions

Language:
English Language

Transcript

00:02

Here’s your shmoop du jour, brought to you by the square root symbol.

00:06

Not to be confused with the square root beer symbol.

00:12

If a = -64 and b = 64, which of the following are not real numbers?

00:19

And here are the potential answers…

00:23

Okay, so we know that a is -64, and b is 64.

00:28

First off, we can substitute the values for a and b.

00:31

The first choice becomes the square root of -64.

00:34

The second becomes the square root of 64, the third becomes the cube root of -64, and

00:40

the fourth becomes the cube root of 64. Now, which of these are not real numbers?

00:45

Remember that a real number is something that exists on a number line.

00:48

That includes natural numbers, integers, and rational numbers.

00:53

What it doesn’t include is imaginary numbers. We get imaginary numbers when we try to take

00:58

an even root of a negative number.

01:04

If we look at option 1, it looks like we’re trying to take the square root of negative 64.

01:10

Well it's imaginary number, since no number squared can equal negative 64.

01:16

Which means option 1 is not a real number.

01:18

Roman numeral 2 is the square root of 64. Well… that’s easy. It’s just 8.

01:23

8 falls within the category of integer, which is a real number.

01:28

Option 3 is the cube root of -64.

01:31

Even though we’re taking the root of a negative number, this number does exists.

01:35

In fact, the cube root of -64 is just negative 4.

01:39

Finally, the cube root of 64.

01:41

The cube root of 64 is, in fact, 4. Choices 2, 3 and 4 all exist.

01:46

However, option number 1 is not a real number.

01:48

Our answer is (A).

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