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June 25th, 2019, 07:49 PM  #1 
Newbie Joined: Jun 2019 From: Cape Town Posts: 2 Thanks: 0  I went awry somewhere with this problem Hi there, I was expected an answer of 2 for this problem, but it seems the correct answer is 8! Where did I go wrong? I replaced the m or 4, then went 42 which got me to 2. Last edited by skipjack; June 26th, 2019 at 06:40 PM. 
June 25th, 2019, 08:05 PM  #2 
Global Moderator Joined: Oct 2008 From: London, Ontario, Canada  The Forest City Posts: 7,950 Thanks: 1141 Math Focus: Elementary mathematics and beyond 
Problem statement: Given $g(m)=\sqrt{m4}$, solve $g(m)=2$. Solution: $\displaystyle 2=\sqrt{m4} \\ 2^2=(\sqrt{m4})^2 \\ 4=m4 \\ 4+4=m \\ 8=m$ 
June 25th, 2019, 08:45 PM  #3 
Newbie Joined: Jun 2019 From: Cape Town Posts: 2 Thanks: 0 
Hi Greg, thanks so much for the solution: I am super confused though  must be missing something fundamental in my math knowledge. For example: In the second line of the solution you added the exponent to both sides of the equation  why? How is that the correct thing to do? Also it feels to me like the root sign disappeared and nothing was done with it but I am sure that's not the case? Yikes, I got a lot to learn Thanks in advance, 
June 25th, 2019, 09:20 PM  #4  
Member Joined: Oct 2018 From: USA Posts: 87 Thanks: 59 Math Focus: Algebraic Geometry  Quote:
Let $a=b$, then you can multiply by $a$ on both sides to get $a^2 = ab$, or you can multiply by $b$ on both sides to get $ab=b^2$. Notice that both $a^2 = ab$ and $b^2 = ab$, so we have $a^2 = b^2$ You can generally apply exponents to both sides like you would add/multiply on both sides. Quote:
So you can get rid of the radical by squaring it. $\sqrt{a}^{2} = (a^{\frac{1}{2}})^{2} = a^{\frac{2}{2}} = a^1 = a$ Last edited by Greens; June 25th, 2019 at 09:24 PM.  
June 26th, 2019, 06:47 PM  #5 
Global Moderator Joined: Dec 2006 Posts: 20,819 Thanks: 2158  
June 27th, 2019, 10:09 AM  #6  
Member Joined: Jun 2019 From: USA Posts: 69 Thanks: 27  Quote:
Philosophy of tackling algebra problems: If you're solving for x, you want to get x on its own on one side of the equation. You generally do this by seeing what was done to x, and then undoing it.* If you have something under a square root, try squaring it. If you have x + 3, try subtracting 3. I think Greg and Greens explained the rest of it quite nicely. *Note, there are other rules and special cases you have to watch out for, but it sounds like you're probably not there yet.  