## invisadude 3 years ago square root problem

1. cathyangs

which is?

the expression $\sqrt[4]{16x ^{2}y ^{7}?}$ is equivalent to which? 2x^(1/2)y^(7/4) , 4x^(1/2)y^(7/4), 2x^8y^28, or 4x^8y^28.

sorry it took me so long

4. BlingBlong

Use the the fundamental rule with radicals that (ab)^(1/2) = (a)^(1/2) * (b)^(1/2)

5. BlingBlong

Do you understand radicals? They are simply used to determine what two (or 4 in this case) identical numbers multiply together to give what is under the radical for example 3^(2) = 9 3*3 = 9 (9)^(1/2) =3

i know but how does that help me find what the equation is equivalent to?

7. BlingBlong

so split up the terms of your problem into (16)^(1/4) * (x^(2))^(1/4) * (y^(7))^(1/4) I will solve the first term to show you how to solve this so we have (16)^(1/4) if you raise 2^(4) you realize you can get 16 so it simplifies to just 2 so now you are left with 2 * (x^(2))^(1/4) * (y^(7))^(1/4) Now to solve a variable what times what give y^(7) that will allow it to be simplified (when multiplying numbers with exponents we add the exponents). so we have 2 * (x^(2))^(1/4) * (y^(4))^(1/4) * (y^(3))^(1/4)

8. BlingBlong

do you know what the next step is?

9. BlingBlong

??

10. BlingBlong

remember the rule that (x^(4))^(1/4) is the same as x^((x/1)(1/4)) = x^(1) = x

um do i i change (x^2)^1/4 into x?

12. BlingBlong

wait apply the rule I just showed you so (x^(2))^(1/4) = x^((2/1)(1/4))

13. BlingBlong

you know how to multiply fractions right and that all numbers can be expressed with a 1 in the denominator

14. BlingBlong

and how to simplify fractions right? (2/4) = (1/2)

ya so um it turns out to be x^1/2?

16. BlingBlong

yup

17. BlingBlong

what is (y^(3))^(1/4) simplified?

2x^1/2 and y^7/4

19. BlingBlong

y^(7/4) can be simplified more you would know that if you read what I posted

20. BlingBlong

you can split up numbers and variables to aid in simplification of them

ya but that is what it looks like as one of the answer choices

22. BlingBlong

oh ok weird sorry for the accusation hope I was helpful