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 Algebra Pre-Algebra and Basic Algebra Math Forum

 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 4-2 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: 7,963 Thanks: 1148 Math Focus: Elementary mathematics and beyond Problem statement: Given $g(m)=\sqrt{m-4}$, solve $g(m)=2$. Solution: $\displaystyle 2=\sqrt{m-4} \\ 2^2=(\sqrt{m-4})^2 \\ 4=m-4 \\ 4+4=m \\ 8=m$ Thanks from topsquark 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
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Quote:
 Originally Posted by hb7of9 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?
Here's why that works.

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:
 Originally Posted by hb7of9 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?
A square root is actually just an exponent of $\frac{1}{2}$. For example $\sqrt{a} = a^{\frac{1}{2}}$

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
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Quote:
 Originally Posted by hb7of9 I replaced the m or 4, then went 4-2 which got me to 2.
Replaced with what and why? What does "went 4-2" mean and why did you do it? June 27th, 2019, 10:09 AM   #6
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Quote:
 Originally Posted by hb7of9 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?
First rule of algebra: Whatever you do to one side of the equation, you have to do to the other.

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. Tags awry, problem Thread Tools Show Printable Version Email this Page Display Modes Linear Mode Switch to Hybrid Mode Switch to Threaded Mode

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