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(3x-1)2 = 12
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(3x-1)(3x-1) = 12 Try toSolve
is that squared?
i guess so.
yes
oh then you have a choice. you can multiply out first, collect terms, set equal zero and solve or just bust through to the answer. your choice
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all the problems i have are squared
\[(3x-1)^2=12\] \[3x-1=\pm\sqrt{12}=\pm2\sqrt{3}\] \[3x=1\pm2\sqrt{3}\] \[x=\frac{1\pm2\sqrt{3}}{3}\] is method two
(x-4)2 + -5, this problem is also squared
you lost me. is that an equal sign?
yes it is suppose to be equal
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\[(x-4)^2=-5\]?
is that it?
yes
ok if you are working with real numbers there is no solution, since the left hand side is a perfect square and so always bigger than zero, but the right hand side is negative. so no real solution
okay
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