Question

4Square root x^20y^28

Answer 1

I... I assume you're being asked to simplify this expression? If so, you should think of the square root as a one-half power (I could explain why that makes sense if you want.) and then simplify the exponents. If I read it like, it would go like this:

4 * sqrt(x^20*y^28) = 4 * (x^20 * y^28)^(1/2)
= 4 * (x^20)^(1/2) * (y^28)^(1/2)
= 4 * x^(20/2) * y^(28/2)
= 4 * x^10 * y^14

Answer 2

no I know how to solve it with a normal 4 in front of the square root but in this case its a tiny 4

Answer 3

A "tiny" four? Is it in the cup on the root operator? If so then it's a "fourth root" instead of a "square root". Fourth roots are "one-quarter" powers.

Answer 4

yeah its the 4th root sorry i forgot what it was called

Answer 5

So it's more or less the same process:

\[\sqrt[4]{x^{20}y^{28}} = (x^{20}y^{28})^{(1/4)}\]

...and then simplify from there.

Answer 6

yeah im having trouble with the simplifying part

Answer 7

Use exponent rules \[\Large (x^a)^b=x^{a*b}\]

Answer 8

yes exactly just multiplie the 20 by 1/4 and the 28 by 1/4

ok. ?

Answer 9

oh so i just multiply?

Answer 10

The exponents, yes.

Answer 11

so it would be 20^5y^7?

Answer 12

Why would it be 20^5?

Answer 13

because 20(1/4) is 5

Answer 14

\[\Large (x^{20}y^{28})^{1/4}=\]

Answer 15

\[x ^{5}y ^{7}\]

Answer 16

Much better.

Answer 17

so is that the answer?

Answer 18

Well, do you see anything else you can do to it?

Answer 19

no

Answer 20

Exactly.