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Precalculus
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Solve by extracting the square roots: 3(3x-1)^2=21
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divide by 3 to get \((3x-1)^2=7\)
then take the square roots
3x-1=\[\sqrt{7}\]? Like this?
yeah don't forget \[3x-1=-\sqrt7\]
how come it's negative?
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could be \(\sqrt7\) or could be \(-\sqrt7\)
ohhhh ok :) so would that be the answer? or is there more to it?
yeah you have to solve for \(x\)
add 1 divide by 3
So it would be:\[\frac{ 1\pm \sqrt{7} }{ 3 }\]
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yes
so is that what they mean by extracting the square roots?
i guess
okie dokie :) thank you so much!!!!! :D
yw
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