1. ## Algebra question

Hey all. The root of my problem is actually from a chemistry question I'm doing, but for some reason I can't wrap my head around the algebraic solution to a formula need to solve the question. It's fairly simple, but for some reason I'm stuck. It gives the answer as x = 0.18 which works when you plug it in, but I can't get to it. (Please ignore the spacing above and below the division sign as it won't recognize my spaces for some reason)

(1.0 - x)
-------------------- = 1.0
(0.50 + x) (1.0 + x)

Here's what I've done:

(1.0 - x)
------------------- = 1.0
(.5 + 1.5x + x^2)

(.5 + 1.5x + x^2) = (1.0 - x)

.5x + x^2 = .5

Therefore:

.5 = x^2 + .5x

I tried the quadratic formula but it didn't give me the right answer. Can anybody help? Thanks.

2. Originally Posted by Jools
Hey all. The root of my problem is actually from a chemistry question I'm doing, but for some reason I can't wrap my head around the algebraic solution to a formula need to solve the question. It's fairly simple, but for some reason I'm stuck. It gives the answer as x = 0.18 which works when you plug it in, but I can't get to it. (Please ignore the spacing above and below the division sign as it won't recognize my spaces for some reason)

(1.0 - x)
-------------------- = 1.0
(0.50 + x) (1.0 + x)

Here's what I've done:

(1.0 - x)
------------------- = 1.0
(.5 + 1.5x + x^2)

(.5 + 1.5x + x^2) = (1.0 - x)

.5x + x^2 = .5 Mr F says: This does NOT follow from the previous line. You are meant to ADD x to both sides .... By the way, thankyou for posting your work. It makes it easy to diagnose your trouble.

Therefore:

.5 = x^2 + .5x

I tried the quadratic formula but it didn't give me the right answer. Can anybody help? Thanks.
And once you have fixed your mistake, I'm sure you know that you must re-arrange the equation into the form quadratic = 0 before using the quadratic formula.

3. Originally Posted by Jools
(.5 + 1.5x + x^2) = (1.0 - x)

.5x + x^2 = .5
Your error is between these two lines. For whatever reasons you've subtracted when moving the -x from the right to the left when you should have added it.

EDIT: Mr F is right, far easier to figure out when work has been posted. Just out of interest what topic is this in chem?

4. Thanks for your reply! In looking at it again I did make an error there, but on my paper I did it the right way, and still didn't get the answer from the book. Starting from where I made the error above:

x^2 + 2.5x - .5 = 0

a = 1 b = 2 c = -.5

Using the quadratic formula I come up with:

-2.5 +/- 1.44 which is either - 1.06 or - 3.94

Sorry about the confusion. And thanks again for the help.

P.S. The chemistry topic I am working on is chemical equilibrium. This formula is for finding the concentrations of reactants when volume changes.

5. Then you are using the quadratic formula incorrectly.
$\frac{-b\pm\sqrt{b^2- 4ac}}{2a}$

The first part, $\frac{-b}{2a}$ is $\frac{-2}{2}= -1$, not "-2.5"
The discriminant is $\sqrt{b^2- 4ac}= \sqrt{2^2- 4(1)(-.5)}= \sqrt{4+ 2}= \sqrt{6}$ which is about 2.45, not "1.44".

6. Ok got it. I wasn't including -b as part of the numerator, I was adding/subtracting it to the resolved fraction. Thans for all the help!