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Square Roots in Equations


Date: 06/12/97 at 23:32:10
From: Carrie Carlson
Subject: Algebra

I'm supposed to solve this:

2(x-4)^2 - 3 = 13

Math is my worst subject. Here's what I think:

2(x-4) - 3 = sqrt(13) or    2(x-4) - 3 = sqrt-(13)
2x - 8 - 3 = sqrt(13) or    2x - 8 - 3 = sqrt-(13)
2x - 5 = sqrt(13)     or    2x - 5 = sqrt-(13)
2x = 5 + sqrt(13)     or    2x = 5 + sqrt-(13)

x = 5 + sqrt(13)      or    x = 5 + sqrt-(13)
    -----------                  -----------
         2                           2

Did I do this right, or did I mess up at the beginning with the 
exponent?  The exponent kind of throws me off because I'm not sure if 
I'm allowed to just take it out of there like that before I multiply 
the parentheses.

Also, could you help me with this problem:

3x^2 - 16 = 0

I'm supposed to solve it like the last one, but I'm lost.

Thanks. This is cool that you people would help me. 
Of course I try to do them first.

Carrie


Date: 06/13/97 at 10:33:39
From: Doctor Wilkinson
Subject: Re: Algebra

I think you recognize that what you have done isn't right.  This is 
good; knowing when you are wrong is the first step towards learning to 
get it right. So let's take another look at that equation:

 2(x-4)^2 - 3 = 13

Every equation has certain difficulties that it puts in the way of 
finding the answer. The biggest difficulty that this equation gives us 
is that exponent. It's natural that you should want to get rid of the 
exponent as quickly as possible, but to do that you have to take the 
square root, and now there's a barrier in the way of just taking the 
square root, because whatever you do in an equation you have to do to 
ALL of both sides. If we take the square root as our first step, the 
result is:

 sqrt[2(x-4)^2 - 3] = sqrt(13)

As you can see, this isn't very helpful because all that stuff that 
you're supposed to take the square root of on the lefthand side is 
pretty messy. So that means that we need to do a little more 
rearranging of the equation before we can take any square roots.

Now you can easily take the square root of (x-4)^2 or even of 
2(x-4)^2, but that's not what you have on the left side of the 
equation: you have that -3 term also. So before you can go taking 
square roots, you have to get rid of that -3. You can do that by 
adding 3 to both sides of the equation.  This gives you:

 2(x-4)^2 = 16

Next you probably want to get rid of the factor of 2 on the left side. 
That's easy since you can just divide both sides by 2.  This gives:

 (x-4)^2 = 8

NOW you have something squared all by itself on the left side, and you 
can go ahead and take square roots.  So you get:

 (x-4) = sqrt(8)  OR  (x-4) = -sqrt(8)

This is because there are TWO numbers whose square is 8, one of them 
positive and the other negative (which you seem to already know 
about).

This gives the two solutions for x:

 x = 4 + sqrt(8)  and  x = 4 - sqrt(8)

With this, you should be able to solve the other problem that you 
gave.

-Doctor Wilkinson,  The Math Forum
 Check out our web site!  http://mathforum.org/dr.math/   
    
Associated Topics:
Middle School Algebra

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