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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)
• In the lower right of the video window is a "gear" symbol. Click on it. You can change video speed setting.
• 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)
• when you divide the (4-x^2) out of your equation on both sides it kills the possibility of finding the second solution.