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need clarification on solving this equation
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which?
\[y ^{3}-10y ^{3 \over 2}+16=0\]
I got y=2 and y=8 so far..I'm not sure if I need to change this to a fraction or what?
Try writing the y^(3/2) in another term, suck as x
if you set x = y^(3/2) you will get x^2-10x+16
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ok. Where do I go from here? :/
factor the equation for (x-2) (x-8) so you get x=2 and x=8
and then substitute x = y^(3/2) and then solve
y^(3/2) = 2 y^(3/2) = 8 y = 2^(2/3) y = 4
okk gotcha! thanks!
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