## Fractional Quadratic equation to standard form

Quadratic equations and inequalities, variation equations, function notation, systems of equations, etc.

### Fractional Quadratic equation to standard form

Hi everyone,
I'm practicing my quadratic equations for exams in a few months and this seems to be the only one in the book giving me problems.

Usually I don't have a problem putting them into standard form and factorising them, perhaps I'm forgetting to do something.
This is the equation I'm working on at the moment, the answers that I am suppose to end up with in the book are x=7/3 and x=-1 ( where x is not actually to -1 because one of the original terms is undefined).

3 - 4/(1+x) = 8/(x^2-1)
3 + 4/(x-1) = 8/(x(x-1)) - simplify
(x-1)(3 + 4/(x-1)) = (8/(x(x-1)))(x-1) - multiply both sides by (x-1)
3x -3 +4 = 8/x
x(3x + 1) = (8/x)(x) multiply both sides by x
3x^2 +x = 8
3x^2 +x -8 = 0

This is where i get stuck because i can't factorise it from here. Unless i got it wrong way in the beginning.

please give me a hand.
GeekBoy

Posts: 5
Joined: Tue Feb 25, 2014 5:41 pm

### Re: Fractional Quadratic equation to standard form

GeekBoy wrote:3 - 4/(1+x) = 8/(x^2-1)
3 + 4/(x-1) = 8/(x(x-1)) - simplify

How are you getting that x(x - 1) = x^2 - x is equal to x^2 - 1?
buddy

Posts: 121
Joined: Sun Feb 22, 2009 10:05 pm

### Re: Fractional Quadratic equation to standard form

im reversing the order of the denominator by using -1.

-1(1-x) = x-1
GeekBoy

Posts: 5
Joined: Tue Feb 25, 2014 5:41 pm

### Re: Fractional Quadratic equation to standard form

GeekBoy wrote:im reversing the order of the denominator by using -1.

-1(1-x) = x-1

Im not asking about -1(1-x) equaling to -x+1. Im asking about x(x-1) equaling to x^2-1.
buddy

Posts: 121
Joined: Sun Feb 22, 2009 10:05 pm

### Re: Fractional Quadratic equation to standard form

Oh, ok. I see, I thought i was taking the common factor x out to simplify it.
because x into x would be x^2 and x in - 1 would give me -x not -1.
It should be (x+1)(x-1)
Am I correct ?
I get the right numbers but the signs aren't right. I get x=-7/3 and x=1 instead of x=7/3 and x=-1, which is what my text book gives me.

3 - 4/(1+x) = 8/(x^2-1)
3 + 4/(x-1) = 8/((x=1)(x-1)) - simplify
(x+1)(x-1)(3 + 4/(x-1)) = (8(x+1)(x-1))/(x+1)(x-1) - multiply both sides by (x+1)(x-1)
3x^2 -3 +4x+4 = 8
3x^2 + 4x + 1 = 8
3x^2 + 4x -7 = 0 factors of -21 that when added equal 4 are +7 and -3
3x^2 -3x +7x -7 = 0
3x(x-1)+7(x-1)
(3x+7)(x-1)=0
that gives me x = -7/3 or x = 1

Am I forgetting a rule ?
GeekBoy

Posts: 5
Joined: Tue Feb 25, 2014 5:41 pm

### Re: Fractional Quadratic equation to standard form

GeekBoy wrote:3 - 4/(1+x) = 8/(x^2-1)
3 + 4/(x-1) = 8/((x=1)(x-1)) - simplify

I hadnt see the prob with the 1st denom. doing the minus through reverses a subtraction but it doesnt turn addition into subtraction so -(1+x) cant be (-1+x). Besides you dont need to do anything to reverse addition; just reverse it because order doesnt mnatter in adding.

$3\, +\, \frac{4}{x\, +\, 1}\, =\, \frac{8}{(x\, +\, 1)(x\, -\, 1)}$

This is an eqn so you can multiply thru by the common denom:

$3(x\, +\, 1)(x\, -\, 1)\, +\, 4(x\, -\, 1)\, =\, 8$
buddy

Posts: 121
Joined: Sun Feb 22, 2009 10:05 pm

### Re: Fractional Quadratic equation to standard form

That gives me x = 5/3 or x = -3. Not what it should work out to, which is x = 7/3 or x = -1. Why is it that I'm getting the same numbers with different signs ??
GeekBoy

Posts: 5
Joined: Tue Feb 25, 2014 5:41 pm

### Re: Fractional Quadratic equation to standard form

GeekBoy wrote:That gives me x = 5/3 or x = -3. Not what it should work out to, which is x = 7/3 or x = -1. Why is it that I'm getting the same numbers with different signs ??

I get the same thing. Why do you think the other ans are rite? If x=-1 then youre dividing by 0.
buddy

Posts: 121
Joined: Sun Feb 22, 2009 10:05 pm

### Re: Fractional Quadratic equation to standard form

I think it's right because that is what my text book gives the answers as.
If you read the full description in earlier posts, you would see that I did mention that one term would be undefined or divided by zero.
What I am looking for is the method to get those answers.

I'm certain my order of operations is wrong somewhere, I just can't see where.
GeekBoy

Posts: 5
Joined: Tue Feb 25, 2014 5:41 pm

### Re: Fractional Quadratic equation to standard form

GeekBoy wrote:I think it's right because that is what my text book gives the answers as.

But x=-1 cant be rite bc it makes division by 0. If the book says its "the answer" then the book is wrong. Try the other ans:

3-4/(10/3)=8/(40/9)
3-(4/1)(3/10)=(8/1)(9/40)
3-(6/5)=(9/5)
18/5-6/5=9/5
12/5=9/5

It doesnt work either. Books can be wrong! Now check your own ans. Do they work?
buddy

Posts: 121
Joined: Sun Feb 22, 2009 10:05 pm