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m^2=13 I put no solution for this one. Am I correct in doing this?
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No. I thought you had this before except with c instead of m
take the square root of both sides you only get into trouble if you take the square root of a negative number. 13 is not negative, so you are ok.
No I never had a problem like this yet
√m √13*1?
√m √3.606
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\[ m^2=13 \] \[ \sqrt{m^2} = ±\sqrt{13} \] \[ m = ±\sqrt{13} \]
ahhhhh damn it!
Thank you again
or if you like \[ \sqrt{ m \cdot m } = ± \sqrt{13} \] \[ m = ± \sqrt{13} \]
Thank you phi
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So it m=√13, and m=-√13 right
@treydwg -----> Yes, the answer you posted is correct. >> So it m=√13, and m=-√13 right
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