In Exercises 59–76, find the indicated root, or state that the ex...

Question

In Exercises 59–76, find the indicated root, or state that the expression is not a real number.___⁶√−1

Accepted Answer

Hey, everyone in this problem, we're asked to determine the indicated root in the given expression. And the expression we're given is the 26th route of -1. We have four answer choices. Option, a negative one, option B zero, option C one and option D not a real number. We're gonna start by writing what we have, we have the 26th root of -1. Now recognize that this is an even route of a negative number. When we have an even root of a negative number, we are always going to get the answer is not a real number. Now, we know that this is true for square roots. OK? If we have the square root of a negative number, we know that that's not a real number and that extends to any even route. Now, if you didn't remember this rule and you were given this question, what could you do? Well, you know the rule for a square root. So we can relate the route we're given to a square root, then we can also find that this is not a real number. So recall, we can write the 26th route of - As -1 to the exponent one divided by 26. And we have this rule that allows us to relate the exponents to radicals or roots. Now, let's also remember our power rule. If we have a, to the exponent M multiplied by N, we can write this as a to the exponent M to the exponent end. OK. So if we have an exponent where something is multiplied together, we can break this up into an exponent of an exponent. Now we have an exponent of one divided by 26, we want to write this as two things multiplied together so that we can break it up. We want one of those things to be one ha and why do we want that? Well, we want that. So we can relate it to a square root. We know that a square rib is equivalent to an exponent of one half. So we can write one divided by 26 as one half multiplied by one divided by 13. So we can break up our exponent and write it as negative one to the exponent, one half all to the exponent one divided by 13. Now the exponent 1/2 we know is equivalent to a square root. We know that the square root of a negative number is going to be not a real number. So inside our brackets, we have not a real number if we have not a real number. And then we take the 13th root of it. It's still going to be not a real number. So the solution here is that we get not a real number which corresponds with answer choice. D. Thanks everyone for watching. I hope this video helped see you in the next one.

In Exercises 59–76, find the indicated root, or state that the expression is not a real number.___⁶√−1

Key Concepts

Roots of Numbers

Real and Complex Numbers

Even Roots of Negative Numbers

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