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# Equation with two rational expressions (old example 3)

Sal solves the following equation and eliminates the resulting extraneous solution: (x^2)/(x+2)=4/(x+2). Created by Sal Khan and Monterey Institute for Technology and Education.

## Want to join the conversation?

• At , couldn't you just use square roots to solve it instead of factoring a quadratic equation?
• when you do sq in the calculator you only get the positive half of the root, you need to remember it could also equal the negative number.

bottom line: you could. :)
• Why is x=-2 not a possibility? even though 4/0 is 0 it is still zero. I understand that is different but I need to organize this better in my mind. Can someone please explain better?
• 4/0 is equivalent to trying to divide 4 by 0, which is undefined, and hence why you cannot have it as a solution.
(1 vote)
• This may sound a silly question but something just occured to me. At first I used a different method to solve this.
(x^2 / (x+2)) - (4 / (x+2)) = 0
(x^2 - 4) / (x+2) = 0
(x+2)(x-2) / (x + 2) = 0
x - 2 = 0
x = 2

I was just wondering, if you can somehow reduce an equation like this to another equation with only one solution, would be safe to assume that said solution is non extraneous?

Thanks.
• The only thing to be wary of is when you cancel the (x+2) factor from the top and bottom. It was okay for this problem, and you got the right answer.
But you must be careful. Let's say that after you cancelled, you were left with (x-2)*(x+2)=0. You would think that the solutions were x=2 and x=-2. However, if you plug -2 into the original problem, your answer would be indeterminate.
The safest thing to do is to append a note x≠-2 after you cancel the (x+2) factor.
• Is there a way to book mark any video ?
• You can just bookmark the page in your browser. Make sure that you are on the page of the video by clicking the video name on top of the frame (if you see it) :)
(1 vote)
• I understand how you can get an extraneous root from difference of perfect squares since the square root of something can be positive or negative. But how come you can get an extraneous solution from a different type of quadratic equation? How can you get an answer from an equation that isn't an answer? If the number is coming from the equation shouldn't it be a solution to it?
• The extraneous root here is also "coming from the equation". Factoring the difference of perfect squares is just how you happen to solve it. With any equation, you simplify it (like when we got rid of the denominators in this case) and then either get a solution/solutions or not. It doesn't matter how you got these solutions, but if you input some "root" into the original equation and it causes you to perform some invalid operation—like division by zero—it's not a proper root, the equation breaks down with it.

It happens because as you go about simplifying the original equation, the information about values that are just out of consideration is lost or they become less obvious. Also, extraneous roots aren't connected to roots that come in pairs (You said, "I understand... ...since the square root can be positive or negative.") One can probably imagine some more exotic case where you get a single root and it's an extraneous one; OR where you get two extraneous roots.
• At Sal says that x cannot be -2 because it would make both terms undefined.
But, I still don't get it:

(-2)^2/(-2+2) = 4/(-2+2)
4/0 = 4/0

Why it isn't a solution ? Yeah, we don't know what is the quotient of 4/0, but, isn't it logical to assume that whatever the quotient is, it is going to be equal to 4/0? I mean, they're both the same thing. (4/0 = 4/0)
• 4/0 is undefined (in a way, its not even a number), so it can't be a solution and if you try to solve it with 4/0 your equation will just be messed up. There are contexts where it is defined, but they don't usually appear until much later in your math studies.
• I`d like to receive an explanation for some other way of solving the problem. Assume that we do not multiply both sides by (x+2), but rather move 4/(x+2) to the left side, so we get (x^2-4)/(x+2) = 0. We can factor x^2-4 as a difference of squares to receive (x+2)(x-2) and then cancel out x+2 in both nominator and denominator, setting a domain of all real numbers except x=-2. We`ll get x-2=0, x=2. As a result, there is NO extraneous solution, there is only one (x=2), while the other one appears to be a restriction, not an extraneous solution.
Can you explain me this? Is it okay that solving an equation the way I propose you fail to find an extraneous solution, instead of this putting a restriction for the domain practically from the start?
• Since there are multiple ways to solve this equation, some of them might not include extraneous solutions. Your other way of solving the equation is just as correct and okay as Sal's way even though it does not show any extraneous solutions.

I hope this helps!
• In the earlier video Rational equations, Sal discusses excluded values. The excluded value is excluded as it would make the denominator 0. Is this the same thing as the extraneous solution? The words are interchangeable? Thanks!
• Not exactly. Excluded values are values that are not in the domain - values that would not be okay to "plug in" because they would cause us to have a 0 in the bottom of a fraction. Extraneous solutions are results that you might get from your work that need to be thrown away because they happen to be one of the excluded values.

It's a bit of a misnomer because "extraneous solutions" are not really solutions at all.
• At around , couldn't you take the (positive and negative) square roots of both sides?
• Why is negative 2 not a solution when 2 is? They both give the same answer. Why is 2 not an extraneous solution then?
(1 vote)
• Let's do the checks for each value and see which one(s) work. That might make it clearer.
The equation is: X^2 / (X+2) = 4 / (X+2)

Here is the check for X=2:
2^2 / (2+2) = 4 / (2+2)
4 / 4 = 4 / 4
1 = 1 The 2 sides match and neither side is undefined. So X=2 is an acceptable solution.

Here is the check for X = -2
(-2)^2 / (-1+2) = 4 / (-1+2)
4 / 0 = 4 / 0
While the 2 sides are the same, they both have division by 0. This is undefined. So, X = -2, is not an acceptable solution. It is considered extraneous because it creates these undefined states which are not allowed.

Hope this helps.