In Exercises 39–64, rationalize each denominator.

10
-----
³√5

Question

In Exercises 39–64, rationalize each denominator.

10
-----
³√5

Accepted Answer

Hello, everyone. We've been asked to eliminate the radical in the denominator by rationalizing the expression we are given is 27 over the cube brew of three, we have four answer choices, all of which have a Q group and none of which have a denominator. So that makes me think our denominator is gonna cancel out. The first thing I'm gonna do is rewrite my original expression. So 27 over the cube root of three, I recall that to rationalize this denominator, I need to multiply numerator and denominator by a value that will cancel out the radical in the denominator. So we are gonna multiply both the numerator and denominator by the cubed root of something to decide what that something is. I need to think, what do I need to multiply three to the first power by to make it three rays to the third power. Since I recall that when I multiply bases with exponents, I add the exponents. I know that I need to multiply by three squared to get the three to the first power to become a three to the third power. So I'm going to multiply numerator and denominator by the third root or the cubed root of three squared. So in doing so I'm gonna multiply straight across my numerator 27 multiplied by the cube root of three squared. In my denominator, I will multiply straight across. Also, I keep the cubed root three to the first power multiplied by three squared will give me three cubed. So my denominator is the cube root of three cubed. And that's exactly what we wanted. We want the exponent and the index to match my numerator. I'm gonna rewrite it and I'm actually gonna write three squared as nine because I don't think it'll hurt us in the long run. So rewriting my, my N reader as 27 multiplied by the cubed root of nine over my denominator. So the cubed root of three cubed, a cube and a cube route will cancel out leaving us with a three in my denominator. So at this step, I have 27 multiplied by the cube root of 9/3. I know I can reduce this one more time because I can simplify 27/3, three will go evenly into both of them. So 27 divided by three is nine. So my num reader is now nine multiplied by the third route of nine. And then my denominator, the three divided by three is one. And I recall that any value in a fraction over one is the same as saying it does not have a denominator So nine multiplied by the cubed root of nine is our solution. And that is answer choice C have a nice day.

In Exercises 39–64, rationalize each denominator. 10 ----- ³√5

Key Concepts

Rationalizing the Denominator

Cube Roots

Multiplying by the Conjugate

Watch next