Question

Adding/subtracting rational expressions question. I know i have to do one of two things

Answer 1

ill write the problem on equation

Answer 2

\[\frac{ x^2+2x }{ 12x+54 }-\frac{ 3-x }{ 8x+36 }\] would i simpligy this to
\[\frac{ x^2+2x }{ 6(2x+9) }+\frac{ -3+x }{ 4(2x+9) }\] or would i mulitply the two equations with the opposite denominators?

Answer 3

You've made a good first step.Remember, to be able to add or subtract two fractions together they need to have the same denominator.

At the moment they're pretty close, the left fraction has 6(2x+9) and the right has 4(2x+9).
The only thing you have to do now is multiply the left fraction by 4 (on top and bottom) and the right fraction by 6 (on top and bottom) to make them both have denominator 24(2x+9)

Does this help?

Answer 4

wolfram alpha says the whole simplified thing should be \[\frac{ x-1 }{ 12 }\]

Answer 5

Work it out after you do the step I said and let me know what you get. I expect that you will be able to factor the numerator and do some cancelling.

Answer 6

Sorry what I said can be simplified a little bit, multiply the left fraction by 2 on top and bottom and the right one by 3 on top and bottom so that they both have a denominator of 12(2x+9).
My bad :)

Answer 7

ok so i got this \[\frac{ 4x^2+14x-18 }{ 24(2x+9) }\] which is \[\frac{ 2(2x^2+7x-9) }{ 24(2x+9) }\]

Answer 8

and thats [(2x^2+7x-9)/12(2x+9)] i dont know how to factor the trinomial cuz it has 2x^2 in the beginning

Answer 9

\[\frac{ 2x^2+7x-9 }{ 12(2x+9) }\]

Answer 10

Good that's what I have.
Do you know how to use the quadratic formula to find the roots of a quadratic equation?

Answer 11

um i know the quadratic formula but i dont think so no

Answer 12

Ok well there is another choice, you could hope that the question is being nice and (2x+9) factors into the numerator, in which case you would then be able to simplify it.

Answer 13

is it the thing where you take a b and c and plugin to b^2-4ac

Answer 14

So rewrite 2x^2 +7x -9 as:
(2x+9)(x+a)=2x^2+9x+2ax+9a=2x^2+(9+2a)x+9a
So you can see that 9+2a = 7 and 9a = -9.
Can you therefore see what your value of a will be, and do you understand what i've done here?

Answer 15

no i have no idea i thought that (2x+9)(x+a) is 2x^2+7x-9 but after you distribute is some crazy polynomial

Answer 16

Ok, what I'm trying to do is to factorise 2x^2 +7x -9 with (2x+9) as one of the factors, because then we would be able to cancel the (2x+9) on the numerator and the denominator, right?

Answer 17

So if we want to have (2x+9) as one of the brackets, then the other brackets will look like (x+a) where a is just a number that we haven't found yet.
So if we are rewriting 2x^2+7x-9 as (2x+9)(x+a), then we can expand the right hand side brackets to find what the value of our a will be. Does this make it a bit clearer?
If you don't like this method you may not have covered enough of quadratic equations to be comfortable enough with them yet. If so, I can just describe how to do it using the quadratic formula if you want? (However, if you did it this way it would impress your teacher!)

Answer 18

Well i have a bunch of examples here Ive written down but the only one that has binomials for the denominators are very simple is x+5 and x-1 so they multiply the two ya know like we did with 4 and 6? and the problems solution ends with (x-1)(x+5) as the denominator. Could I just multiply the two denominators? or would that not work?

I really dont understand what youve written so I would like you to show me the quadratic formula way but I havent had to use that in the lessons section for adding/subtracting rational expressions

Answer 19

Ok that's not a problem. If you multiplied the denominators together it would only complicate things really, you've been given 2 denominators which are already fairly similar.

Onto the quadratic equation: it finds the roots of a quadratic equations which has the form
\[ax^2 +bx+c=0\]
The roots are given by:
\[x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}\]

Now, the quadratic equation we have is:
\[2x^2+7x-9\]
Are you able to tell me the roots to that?

After you've done that, are you comfortable with the fact that, once factorised, the equation will look like this:
\[(x-r)(x-s)\]
Where r and s are the roots of the quadratic that we found?

Answer 20

Typo: the quadratic formula finds the roots of a quadratic equation.

Answer 21

well I put in the problem at wolframalpha.com and it says that the root is x=1

Answer 22

Yes one of the roots is 1, can you find the other one?

Answer 23

well when i do it myself i get \[\frac{ -7\pm \sqrt{-23} }{ 4 }\]

Answer 24

Hmm that doesn't look right. Your b^2-4ac should be:
\[49-4\times2\times(-9)=49-8\times(-9)=49+72=121=11^2\]

Answer 25

(Be very careful with the minus signs!)

Answer 26

oh yeah sorry, ok give me one second

Answer 27

No worries, take your time.

Answer 28

what do i do after \[\frac{ -7\pm \sqrt{11} }{ 4 }\] ?

Answer 29

That's not quite right, it should be:
\[\frac{-7\pm\sqrt{11^2}}{4}=\frac{-7\pm11}{4}\]
This gives you your 2 roots. You can see that taking the plus sign gives you:
\[\frac{-7+11}{4}=\frac{4}{4}=1\]
Can you tell me the other root?

Answer 30

oh ok so its \[\frac{ -7\pm11 }{ 4 }\] so you do the plusminus and thats -7-11=(-17/4) or -4 1/4 ?

Answer 31

\[\frac{-7-11}{4}\]
Is not -17/4, try again with the numerator :)

Answer 32

i havent sleept cuz i was catching for a test i have to do today so its a little hard to focus xD

-18/4 or -4 1/2?

Answer 33

please tell thats not wrong

Answer 34

Ha don't worry we've all been there, I'll try and help out as much as I can.
Yes, -18/4 is correct. The other one should just be 1 though, we already figured that out.
-18/4 cancels to -9/2, agree? So we now have our two roots. Tell me if there's anything you don't understand here first before we move on to the last step.

Answer 35

well isnt 18/4 turned into a mixed number by dividing 18 by 4 getting 4 remainder 2 and that goes over 4 and then simplified to 1/2?

Answer 36

If you have the fraction 18/4 you can notice that both the numerator and denominator are multiples of 2, so divide top and bottom by 2 to get 9/2.

Answer 37

\[\frac{ 2(x ^{2} +2x) - 3(3-x) }{12(2x+9)} = \frac{ 2x^{2} +4x +3x -9 }{ 12(2x+9)} = \frac{ (2x+9)(x-1) }{ 12(2x+9) }\]

Answer 38

Now, when you factorise a quadratic, you put it in the form:
\[(x-r)(x-s)\]
Where r and s are the roots of the quadratic. So, our roots are -9/2 and 1, so we put these in for r and s:
\[(x-\frac{-9}{2})(x-1)\]
Two minus signs make a positive, so we now have:
\[(x+\frac{9}{2})(x-1)\]
Now, you can multiply all terms in the first brackets by 2 to get:
\[(2x+9)(x-1)\]
And now you are in a position to cancel the (2x+9) with the denominator.
Please tell me if there's anything I can explain any better.

Answer 39

you guys are getting a little side-tracked I think...

Answer 40

No Algebraic. He was having trouble factorising the numerator so I'm helping him.

Answer 41

i dont think i understand the quadratic way either :(

Answer 42

What way have you been taught to factorise in school?

Answer 43

If you tell me the way you've been learning in school/college then I can describe how to do it the way that makes sense to you the most.

Answer 44

you find a shared factor and then you take it out put parenthesis after it and factor, like divide what ever was there to leave there

Answer 45

My apologies, it seems that I'm having some connection problems.
So have you not done any factorisation which looks similar to:
x^2+2x+1 = (x+1)(x+1)?