Question

√(9y4)+12√(y2)-3√(y6)-√(4y4)

Answer 1

Are you trying to simplify?

Answer 2

find the exact value

Answer 3

Are the variables to a power. Ex. 9y^4.....y^2 ?

Answer 4

yea. sorry bout that.

Answer 5

So, you are solving for y?

Answer 6

So I got 3 real y values. and 4 imaginary.

Answer 7

these are the answers:
8y2 + 9y3

11y2

11y5

2y2 + 9y3

Answer 8

I didn't get any of those.

Answer 9

ugh. stupid math.

Answer 10

I have the answer. Are looking for a simplified version, or for the zeros?

Answer 11

You still there?

Answer 12

simplified

Answer 13

~((9y^(4)))+12~(y^(2))-3~(y^(6))-~(4y^(4))

Remove the parentheses.
~(9y^(4))+12~(y^(2))-3~(y^(6))-~(4y^(4))

Pull all perfect square roots out from under the radical. In this case, remove the 3y^(2) because it is a perfect square.
3y^(2)+12~(y^(2))-3~(y^(6))-~(4y^(4))

Pull all perfect square roots out from under the radical. In this case, remove the y because it is a perfect square.
3y^(2)+(12*y)-3~(y^(6))-~(4y^(4))

Multiply 12 by y to get 12y.
3y^(2)+(12y)-3~(y^(6))-~(4y^(4))

Remove the parentheses around the expression 12y.
3y^(2)+12y-3~(y^(6))-~(4y^(4))

Pull all perfect square roots out from under the radical. In this case, remove the y^(3) because it is a perfect square.
3y^(2)+12y+(-3*y^(3))-~(4y^(4))

Multiply -3 by y^(3) to get -3y^(3).
3y^(2)+12y+(-3y^(3))-~(4y^(4))

Remove the parentheses around the expression -3y^(3).
3y^(2)+12y-3y^(3)-~(4y^(4))

Pull all perfect square roots out from under the radical. In this case, remove the 2y^(2) because it is a perfect square.
3y^(2)+12y-3y^(3)-2y^(2)

Since 3y^(2) and -2y^(2) are like terms, add -2y^(2) to 3y^(2) to get y^(2).
y^(2)+12y-3y^(3)

Reorder the polynomial y^(2)+12y-3y^(3) alphabetically from left to right, starting with the highest order term.
-3y^(3)+y^(2)+12y

Answer 14

THATS WHAT I GOT TOO!

Answer 15

hmm. Why didn't you post it eariler?

Answer 16

cause i thought it wasn't the right answer

Answer 17

I see....