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# Equations with one rational expression (advanced)

Sal solves the following equation by first simplifying the rational expression: x^2-(x^2-4)/(x-2)=4. Created by Sal Khan and Monterey Institute for Technology and Education.

## Want to join the conversation?

• Are there any practice problems for rational equations? • The interesting thing about this problem is that, if you don't spot the difference of squares and factor out (x-2), as Sal did, and instead multiply through by (x-2) to get it out of the denominator of the second term, you will be left with a third degree polynomial with three roots, one of which happens to be 2, the excluded value. Attempting to solve it this way inadvertently adds an extra root. Factoring out the (x-2), as Sal did, turns the problem into a quadratic, with only two roots. It wasn't immediately clear to me where the extra root "went" or why we needed to worry about excluding it, until I tried to solve it the other way. Hope this helps someone. • This isn't about the video, but I was wondering how to solve this: (2/m+2) - (m+2/m-2) = 7/3 and if there were any videos on this that could help? I have tried solving it myself and ended up with m^2+19m+8 ; which I'm unsure of how to simplify, if it can be simplified, or even if it is correct.
(1 vote) • Here's how I would solve this.
2 / ( m + 2 ) - ( m + 2 ) / ( m - 2 ) = 7 / 3
(Otherwise, after removing parentheses you'd be left with 5 terms on the lefthand side
2/m + 2 - m - 2/m + 2
which would simplify to
4 - m = 7 / 3 which seems too simple at this level.)
Second, once the problem has the correct parentheses, get a common denominator for both sides of the equation. Then set the numerators equal to each other.
Third, get all terms onto the lefthand side of the equation.
Fourth, factor the terms as the product of two binomials.
There will be 2 real roots.
Do let me know how you get on and if I can help you further.
• I have to solve the problem 2/c-2=2-4/c (the denominator c only belongs to 4). I can't solve it because I don't know what the LCD is. How do you figure out the LCD in this problem? • at ...why did he subtract 4 to get a 0 on the one side instead of just adding 2 + 4 so you would end up with x^2 - x = 6?

Wouldn't x^2 - x^1 = just x? Therefore, x=6?

I don't understand how he knew that you would have to get a zero on the one side and then factor it out??

Thanks for any help and clarification!
(1 vote) • He subtracts 4 from both sides because he wants to turn it into a quadratic equation. (ax^2+bx+c). When you see x raised to the 2 power, x^2, you will want to turn it into a quadratic equation that is equal to zero. Once that is done, now you can factor to solve the equation and find the answers.

Or, using the quadratic equation, you can use the quadratic formula to solve the equation, if that is easier than factoring. Quadratic formula: -b+-sqrt b^2-4(a)(c)/2a. Or you could complete the square to solve it. Just know, it needs to be in the form ax^2+bx+c=0 when you have a x^2 term and that there are 3 ways to solve it once you've converted it into a quadratic equation
ax^2+bx+c=0:
Complete the square
Factor
HInt:
x^2 term= turn into quadratic equation equal to zero
ax^2+bx+c=0

If you don't know what I"m talking about, you should watch the videos in the quadratics section of algebra.
• I have a question concerning a problem on Khan Academy that I need help understanding. It goes like this.

36y^2 - 1/ 30y + 5 = 7
Before you even start, you set the denominator as equal to zero, ie.
30y + 5 = 0
We then end up with y = -1/6.
Then we disregard -1/6 as extraneous if it ever is a possible answer. But, -1/6 makes the equation equal and valid. So how can it be extraneous if it makes the equation equal and valid? Is it just an equation that happens to work out that way, or is it something else? I hope that makes sense.
Thanks!
(1 vote) • -1/6 does make the equation `30y+5=0` valid. It is a perfectly correct solution to that equation. The equation we are looking at, however, is not that, but `36y^2 / 30y+5=7`. If one attempts to put -1/6 in as the solution, you would be dividing by 0 (since 30y+5=0), and since dividing by 0 is against the rules (it just doesn't work), -1/6 is an extraneous solution.
• This is more of a suggestion. Sal speaks so fast during this video that I had trouble staying up. Is there a way on here to slow down the playback? Thanks.
(1 vote) • why did he not add 2 to both sides at ?
(1 vote) • At He Lost Me. Can Someone Explain What He Is Talking About?
(1 vote) • I tried to use another way to solve the equation, but I can just get one of the two answers:
(4-x^2)/(x-2)=4-x^2
(4-x^2)/(x-2)=(4-x^2)/1.....(1)
(x-2)(4-x^2)=(4-x^2)*1......(2)left up expression multiply right down, and left down multiply right up.
x-2=1.......................(3)left side and right side divided by (4-x^2) at the same time.
x=3
I could just get one result and I don't know where I went wrong.
Can somebody help me? thanks
(1 vote) ## Video transcript

Solve the equation: x squared minus x squared minus 4 over x minus 2 is equal to 4. And they tell us that x won't or cannot be 2 because if it was 2, then this would be undefined. It would be dividing by 0. So let's see what x is. Let's see the x that actually satisfies this equation. So you might be tempted to try to express this with the common denominator of x minus 2 and then add these two expressions. But the thing that jumps out at me initially is that we have x squared minus 4 on the numerator, which is a difference of squares. Or if you factor it out, it's x plus 2 times x minus 2. So we should be able to factor this x minus 2 out. So let's do that. So if we were to rewrite it, this is equivalent to x squared minus-- instead of writing x squared minus 4, we know that's a difference of squares. That is x plus 2 times x minus 2. All of that over x minus 2. And that is equal to 4. And this whole time we're assuming that x won't be equal to 2. And because x does not equal to 2, x minus 2 divided by x minus 2 is going to be defined. And it will be 1. So those two will cancel out. And so we're left with x squared minus x plus 2 is equal to 4. We can distribute the negative sign. And I'll just arbitrarily switch colors here. We can distribute the negative sign, so we get x squared minus x minus 2 is equal to 4. And what we want to do is put this in the form ax squared plus bx plus c is equal to 0. That allows us to either factor it or apply the quadratic equation or complete the square, or any of the ways that we know how to solve quadratics. So let's do that. Let's get a 0 on the right-hand side. The best way to do that is to subtract 4 from both sides of this equation. Subtract 4 and we are left with x squared minus x. Negative 2 minus 4 is negative 6. And then 4 minus 4 is 0. That was the whole point. So we have x squared minus x minus 6 is equal to 0. Let me write it up here. x squared minus x minus 6 is equal to 0. And this looks factorable. We just have to think of two numbers that when we multiply them give us negative 6. So they're going to have different signs. When I add them I'll get negative 1. So it looks like negative 3 and positive 2 work. So if we do x minus 3 times x plus 2. It's a little bit of trial and error, but 6 doesn't have that many factors to deal with and 3 and 2 are one apart. They have different signs, so that's how you can think of how we get to that conclusion. Negative 3 times 2 is negative 6. Negative 3 plus 2 is negative 1. So that is equal to 0. So we have two possible ways to get 0. Either x minus 3 is equal to 0 or x plus 2 is equal to 0. And then if we take x minus 3 is equal to 0, if we add 3 to both sides of that equation, we get x is equal to 3 or, if we subtract 2 from both sides of this equation, we get x is equal to negative 2. So both of these are solutions. And let's apply them into this equation to make sure that they work. Because these are solutions to essentially, the situation where we got rid of the x minus 2. Maybe it had some type of side effects. So let's just make sure that both of these actually work in the original equation. So let's try x is equal to 3 first. So you get 3 squared minus 3 squared over 4. Sorry, 3 squared minus 4 over 3 minus 2. So this is equal to 9 minus-- 3 squared is 9 --minus 4, which is 5, over 1. So 9 minus 5, which is equal to 4. Which is exactly what we needed it to equal. And let's try it with negative 2. So if I take negative 2 squared and I have a minus negative 2 squared minus 4, all of that over negative 2 minus 2. So negative 2 squared is 4. Minus negative 2 squared, which is 4. Minus 4. All of that over negative 2 minus negative 2 over negative 4. Well 4 minus 4 is 0, so this whole thing is just going to become 0. So this whole thing is going to equal 4. So both of these solutions work. And that makes sense, because when we actually canceled this out, we actually didn't fundamentally change anything about the equation. Only if you had the situation where the x would have been equal to 2. That's the only thing that you're really changing. So that's why it makes sense that both of these solutions work.