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√(9y4)+12√(y2)-3√(y6)-√(4y4)

Mathematics
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Are you trying to simplify?
find the exact value
Are the variables to a power. Ex. 9y^4.....y^2 ?

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Other answers:

yea. sorry bout that.
So, you are solving for y?
So I got 3 real y values. and 4 imaginary.
these are the answers: 8y2 + 9y3 11y2 11y5 2y2 + 9y3
I didn't get any of those.
ugh. stupid math.
I have the answer. Are looking for a simplified version, or for the zeros?
You still there?
simplified
~((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
THATS WHAT I GOT TOO!
hmm. Why didn't you post it eariler?
cause i thought it wasn't the right answer
I see....

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