Help! 5√2y + 3√32y - 2√27y^2?

Question

Hi, please explain how to do this problem. I don't understand how to do this at all and it would be appreciated if you would explain this to me. I need to simplify the square roots as much as possible. Thanks ^^

Yes the variables are under the radical.

Answer 1

5√2y + 3√32y - 2√27y²
= 5√2y + 3 * √16 * √2y - 2 √9y² * √3
= 5√2y + 3 * 4 * √2y - 2 * 3y * √3
= 5√2y + 12√2y - 6y *√3
17√2y - 6y√3
~~~

Answer 2

You can only combine terms with radicals in addition or subtraction if the radicands are all the same. You can always "pull" a perfect square term out from under the radical.

In this case, there isn't much you can do directly with the first term. With the second term, you can look at 32 as 16*2, and 16 being a perfect square, can be pulled out as 4 (4x4=16). Then you have
5 sqrt(2y)+ 3x4 sqrt(2y).......
These terms can be combined to 17 sqrt(2y).
With the last term, you can look at 27y^2 as
3 x (9y^2). 9y^2 is a perfect square (3y x 3y). So you get - 6y sqrt(3).
All together you have 17 sqrt(2y) - 6y sqrt(3).

Answer 3

5√2y + 3√32y - 2√27y^2 =

= 5*√(2y) + 3*√(16*2y) - 2√(9*3*y*y)
= 5*√(2y) + 3*4*√(2y) - 2*3*y√(3)
= (5+12)√(2y) - 6y *√(3)
= 17√(2y) - 6√3 * y
This is the answer.
But I strongly feel that the last term is not y^2 but only y.
in that case I would like to give the answer as well.

5√2y + 3√32y - 2√27y =
= √(25*2y) + √(9*16*2y) - √(4*27y)
taking (√y) as common term from all of them.
=√y * √(50+288-108)
= √(230) *√y
= √(2*5*13) *√y
as there is not perfect square term in the factors of 230, we will keep it as it is.
Hence,
= √(230*y) is the final answer.

Answer 4

I assume variables are under the radicals as well????
5√(2y) + 3√(32y) - 2√(27y^2)
5√(2y) + 3√(16)(2y) - 2√(9y^2)(3)
5√(2y) + 12√(2y) - 6y√(3)
17√(2y) - 6y√(3)