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Mathematics 16 Online
OpenStudy (anonymous):

solve 3y^2-72=0

OpenStudy (anonymous):

\[3y^2-72=0\]

OpenStudy (anonymous):

Sp, we're solving for y.

OpenStudy (anonymous):

So*

OpenStudy (anonymous):

subtract 3y^2 to both sides.

OpenStudy (jdoe0001):

\(\bf 3y^2-72=0\implies 3y^2\cancel{ -72+72 }=0+72\implies 3y^2=72 \\ \quad \\ \cfrac{\cancel{ 3 }y^2}{\cancel{ 3 }}=\cfrac{\cancel{ 72 }}{\cancel{ 3 }}\implies y^2=24 \\ \quad \\ taking\qquad \sqrt{\qquad } \\ \quad \\ \sqrt{y^2}=\sqrt{24}\implies ?\)

OpenStudy (anonymous):

I can't think right now, but your answer would be y=2 sqrt6

OpenStudy (anonymous):

@jdoe0001 will that be correct?

OpenStudy (anonymous):

y = 2 square root 6

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