1. ## Simplify

M = 2 - square_root(u - 3)
u = (v - 2)^2 + 3

Prove that M = v.

2. ## Re: Simplify

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?

3. ## Re: Simplify

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?

4. ## Re: Simplify

Your use of the square root is incorrect. $\displaystyle \sqrt{x^2}= |x|$, NOT x.
So $\displaystyle \sqrt{(v- 2)^2}= |v- 2|$. Whether that is equal to v- 2 or 2- v depends upon whether $\displaystyle v\ge 2$ or $\displaystyle v< 2$.

If you are not told whether v is larger than 2 or not, the best you can say is that
$\displaystyle 2- \sqrt{((v- 2)^2+ 3)- 3}= 2- \sqrt{(v- 2)^2}= 2- |v- 2|$. That is either v (if $\displaystyle v\ge 2$) or 4- v (if $\displaystyle v\le 2$).

5. ## Re: Simplify

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.

6. ## Re: Simplify

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

$M=2-|v-2|$

$M=\begin{cases}v &v\leq 2\\ 4-v &v>2\end{cases}$

7. ## Re: Simplify

I am sorry that I did not mention that it was given v <= 2 which for romsek explanation M = v.

8. ## Re: Simplify

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