M = 2 - square_root(u - 3) u = (v - 2)^2 + 3 Prove that M = v.
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Originally Posted by joshuaa M = 2 - square_root(u - 3) u = (v - 2)^2 + 3 Prove that M = v. $M = 2 - \sqrt{u - 3} \implies \sqrt{u - 3} = 2 - M.$ I wonder what might happen if you now square both sides?
I don't agree this is the convenient way. I prefer to do it this WAY. M = 2 - square_root((v - 2)^2 + 3 - 3) = 2 - v - 2 = - v Why does my answer become -v, Not +v?
Your use of the square root is incorrect. , NOT x. So . Whether that is equal to v- 2 or 2- v depends upon whether or . If you are not told whether v is larger than 2 or not, the best you can say is that . That is either v (if ) or 4- v (if ).
Originally Posted by joshuaa I don't agree this is the convenient way. I prefer to do it this WAY. M = 2 - square_root((v - 2)^2 + 3 - 3) = 2 - v - 2 = - v Why does my answer become -v, Not +v? $\sqrt{(v-2)^2+3-3}=|v-2|$ So I do not agree with the premise of the question.
Originally Posted by joshuaa I don't agree this is the convenient way. I prefer to do it this WAY. M = 2 - square_root((v - 2)^2 + 3 - 3) = 2 - v - 2 = - v Why does my answer become -v, Not +v? The way the square root is defined on real numbers $\sqrt{x^2}=|x|$ so you end up with $M=2-|v-2|$ $M=\begin{cases}v &v\leq 2\\ 4-v &v>2\end{cases}$
I am sorry that I did not mention that it was given v <= 2 which for romsek explanation M = v.
Originally Posted by joshuaa M = 2 - square_root(u - 3) u = (v - 2)^2 + 3 Prove that M = v. I don't think it would be proved..not sure
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