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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

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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