Rules for when you can "do one thing to one side of the equation and the other"

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chromechris

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I was solving the following problem, but am unsure if it is ok to solve it the way I am doing it, because I am afraid that I am changing the content of the answer. Here is the problem I was working on:

[MATH] Solve for v, where v is a real number. [B]√[/B]8v-11 = [B]√[/B]6v+9 What I would do to solve this: ([B]√[/B]8v-11)[SUP]2[/SUP] = ([B]√[/B]6v+9)[SUP]2[/SUP] 8v-11=6v+9 Subtract 6v from both sides 2v-11 = 9 Add 11 to both sides 2v = 20 Divide both sides by 2 v = 10 [/MATH]
Now, I am not sure if it is ok to square both sides like I did in my first step, because like I mentioned earlier, I don't want to change the content of the equation (I'm not sure if I am doing so or not by squaring both sides). I made up a sample problem to help me see if I was doing so, and I think I was, here is the sample problem:

[MATH] Problem: 3 = [B]√9 Solution: 3=3 Now if I square both sides, the equation is true, but the value of it changes: (3)[SUP]2[/SUP] = ([B]√9)[SUP]2[/SUP][/B] 9 = (3)[SUP]2[/SUP] 9 = 9 The value went from 3 to 9[/B] [/MATH]
I guess I am looking for clarification to know if squaring both sides as my first step for the first problem I listed is ok, and why doing so is not changing the content of my answer (I don't know if it is or is not).
 
Could you use notation that shows the _exact_ extent of the root sign? Then we can discuss the question.

Regarding 3=3. I would call this an equality, not an equation. If you square both sides it will remain an equality.
 
The extent of the square root goes over all constant and variables on both sides of the equation. I don't know how to format the full square root, is there any wiki on how to do it?
 
chromechris said:
The extent of the square root goes over all constant and variables on both sides of the equation. I don't know how to format the full square root, is there any wiki on how to do it?
Click to expand...

Start with this:

LaTeX Math Formatting is Available

A system named LaTeX is available in the forum, to generate math symbols and formatting. Using LaTeX requires learning programming codes and syntax. Tutorials and examples are available on the Internet, and some references and links appear below. (As an alternative to LaTeX, the purplemath site...
www.freemathhelp.com

If you are in a hurry, at least use parentheses: √(8v-11) = √(6v+9)
 
You can apply any function to both sides of an equation and the two results will be equal. That is because, by definition, a function, applied to "a" gives a single result not two or more. If a= b then f(a) and f(b) are f applied to the same number (that is what "a= b" means) so give the same result, f(a)= f(b).

Notice that "square root", as used in many elementary algebra texts, so that "square root of 4 is \(\displaystyle \pm 2\)", is NOT a function! Instead, we have to use \(\displaystyle +\sqrt{x}\), so that \(\displaystyle +\sqrt{4}= 2\) and \(\displaystyle -\sqrt{x}\) so that \(\displaystyle -\sqrt{4}= -2\). By convention, in order treat it as a function, \(\displaystyle \sqrt{x}\) is interpreted as \(\displaystyle +\sqrt{x}\).
 
chromechris said:
I was solving the following problem, but am unsure if it is ok to solve it the way I am doing it, because I am afraid that I am changing the content of the answer. Here is the problem I was working on:

Solve for v, where v is a real number.
√(8v-11) = √(6v+9)
...
Now, I am not sure if it is ok to square both sides like I did in my first step, because like I mentioned earlier, I don't want to change the content of the equation (I'm not sure if I am doing so or not by squaring both sides). I made up a sample problem to help me see if I was doing so, and I think I was, here is the sample problem:
...
I guess I am looking for clarification to know if squaring both sides as my first step for the first problem I listed is ok, and why doing so is not changing the content of my answer (I don't know if it is or is not).
Click to expand...
If you apply any function to both sides of an equation, it will remain true if the original was true. (Yes, the values change, but that doesn't affect the equality. So squaring both sides is a valid thing to do ... but:

In your problem, you have the opposite issue: When you apply a non-invertible function, the new equation may be true when the original was not. For example, if you square both sides of 3 = -3, which is false, you get 9 = 9, which is true. Any solution you end up with may be a solution of the wrong equation; this is called an extraneous solution. So in your example, you have to check the solution in the original equation, and discard it if it doesn't work. (It would fail because of a wrong sign somewhere.)
 
Suppose x=-3. There is no argument here at all as x=-3 and no other value. It is given that x=-3.

Now if were to square both sides we will get x2=9.

Now if x2=9 it follows that x= +/- 3. Based on what we know (that x=-3) we reject the solution of x=3 and keep the solution x=-3.

Is this what you are concerned with?
 
Ok, so I guess I mostly just need to check that my answer to a solution gives equality in the equation if I plug it back in, to make sure that it is not an extraneous solution?
 
Yes.

For an equation like yours, where two radicals are equal, I don't think you can get the kind of extraneous solution I mentioned (solving an equation with the wrong sign); but you can have something like this:

[MATH]\sqrt{2x-11} = \sqrt{x-9}[/MATH]​

Try solving that, and see what happens.
 
?? Are you under the impression that \(\displaystyle 2\sqrt{-7}= \sqrt{-7}\)?
If this was in response to Dr. Peterson's example, I believe his point was that, given that \(\displaystyle \sqrt{2x- 11}= \sqrt{x- 9]}\), squaring both sides, \(\displaystyle 2x- 11= x- 9\). Adding 11 to both sides and subtracting x from both sides, x= 2. Putting x= 2 into \(\displaystyle 2x- 11= x- 9\) gives 4- 11= -7= 2- 9, a true statement. But putting it back into the original \(\displaystyle \sqrt{2x-11}= \sqrt{x- 9}\) gives \(\displaystyle \sqrt{-7}= \sqrt{-7}\) which, if we are talking about real numbers, does not make sense.
 
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chromechris said:
x=2
[MATH] \sqrt{-7}=\sqrt{-7} [/MATH]
Click to expand...
HallsofIvy in post #11 is right.

Your work is not wrong; but this kind of problem is generally taught only in the context of real-number algebra, where we say that [MATH]\sqrt{-7}[/MATH] is undefined, because it is not a real number. In that context, the domain of [MATH]\sqrt{x}[/MATH] is [MATH]x\ge 0[/MATH], so that [MATH]x = 2[/MATH] is not a valid solution because [MATH]-7[/MATH] is not in the domain. Your check failed.

My point was that this is a different way in which you can get an extraneous solution to a radical equation, namely due to domain issues. Again, you have to check your answer (as you did), but in this case you have to recognize the domain issue, which I admit is a bit subtle if you didn't come to this problem through a textbook chapter than has mentioned domains.
 
Dr.Peterson said:
HallsofIvy in post #11 is right.

Your work is not wrong; but this kind of problem is generally taught only in the context of real-number algebra, where we say that [MATH]\sqrt{-7}[/MATH] is undefined, because it is not a real number. In that context, the domain of [MATH]\sqrt{x}[/MATH] is [MATH]x\ge 0[/MATH], so that [MATH]x = 2[/MATH] is not a valid solution because [MATH]-7[/MATH] is not in the domain. Your check failed.

My point was that this is a different way in which you can get an extraneous solution to a radical equation, namely due to domain issues. Again, you have to check your answer (as you did), but in this case you have to recognize the domain issue, which I admit is a bit subtle if you didn't come to this problem through a textbook chapter than has mentioned domains.
Click to expand...

So when you mention domain, you are referring to the equation in the context of a coordinate graph right?
 
Not really; I'm referring to the domain of a function. Graphs aren't really part of the problem. But it's true that when we graph on a plane, we assume real numbers only.
 
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