My Math Forum  

Go Back   My Math Forum > High School Math Forum > Algebra

Algebra Pre-Algebra and Basic Algebra Math Forum


Thanks Tree3Thanks
  • 1 Post By greg1313
  • 2 Post By Greens
Reply
 
LinkBack Thread Tools Display Modes
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.
hb7of9 is offline  
 
June 25th, 2019, 08:05 PM   #2
Global Moderator
 
greg1313's Avatar
 
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
greg1313 is offline  
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,


hb7of9 is offline  
June 25th, 2019, 09:20 PM   #4
Member
 
Greens's Avatar
 
Joined: Oct 2018
From: USA

Posts: 90
Thanks: 61

Math Focus: Algebraic Geometry
Quote:
Originally Posted by hb7of9 View Post
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 View Post
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. $\sqrt{a}^{2} = (a^{\frac{1}{2}})^{2} = a^{\frac{2}{2}} = a^1 = a$
Thanks from greg1313 and topsquark

Last edited by Greens; June 25th, 2019 at 09:24 PM.
Greens is offline  
June 26th, 2019, 06:47 PM   #5
Global Moderator
 
Joined: Dec 2006

Posts: 20,969
Thanks: 2218

Quote:
Originally Posted by hb7of9 View Post
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?
skipjack is offline  
June 27th, 2019, 10:09 AM   #6
Senior Member
 
Joined: Jun 2019
From: USA

Posts: 196
Thanks: 78

Quote:
Originally Posted by hb7of9 View Post
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.
DarnItJimImAnEngineer is online now  
Reply

  My Math Forum > High School Math Forum > Algebra

Tags
awry, problem



Thread Tools
Display Modes






Copyright © 2019 My Math Forum. All rights reserved.