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Mathematics 23 Online
OpenStudy (goldps):

Find y y^(3/2)=15y

OpenStudy (anonymous):

\[\begin{align*} y^{3/2}&=15y\\ y^{3/2}-15y&=0\\ y^{3/2}-15y^{2/2}&=0\\ y^{2/2}\left(y^{1/2}-1\right)&=0\\ y\left(\sqrt y-1\right)&=0 \end{align*}\]

OpenStudy (goldps):

ok, not sure what comes after that

OpenStudy (triciaal):

check your factorization

OpenStudy (anonymous):

\[\begin{align*} y^{3/2}&=15y\\ y^{3/2}-15y&=0\\ y^{3/2}-15y^{2/2}&=0\\ y^{2/2}\left(y^{1/2}-\color{red}{15}\right)&=0\\ y\left(\sqrt y-\color{red}{15}\right)&=0 \end{align*}\] @triciaal thanks :)

OpenStudy (triciaal):

welcome y = 0 or (rt y - 15 ) = 0 rt y = 15 (rt y)^2 = 15^2 y = 225 solution y = 0, 225

OpenStudy (triciaal):

need to check in original

OpenStudy (goldps):

nice, that was correct. Now if i can just remember how to do that on the test, i'll be golden

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