Partial Fractions with Ugly Coefficients

[AntSC]

Homework Statement

The question is stated at the top of the attached picture with a solution
20160303_095831.jpg
The correct results of the coefficients are A=2, B=-5, C=1
I have tried this problem multiple times and am still getting ugly coefficients. I have no idea why. A fresh pair of eyes would be welcome.
I have even had mathematica solve it for me to get some insight but it solves using matrices, which is not helpful when attempting it algebraically.

 

[SteamKing]

Homework Statement

The question is stated at the top of the attached picture with a solution
20160303_095831.jpg
The correct results of the coefficients are A=2, B=-5, C=1
I have tried this problem multiple times and am still getting ugly coefficients. I have no idea why. A fresh pair of eyes would be welcome.
I have even had mathematica solve it for me to get some insight but it solves using matrices, which is not helpful when attempting it algebraically.

The equation you wrote at the top:

97x + 35 ≡ A(5x + 2)² + B(2x-3)(5x+2) + C(2x-3)

is correct. The rest of your work below that is what's incorrect.

The idea is to expand the terms of the equation above and equate like with like, not to solve the individual factors like you appear to have done.
In other words, setting x = -2/5 and then doing some manipulation, where did that come from?

After doing the expansions, you should have three equations each in the three unknowns A, B, and C to solve for the values of the unknown coefficients.

 

[BvU]

setting x = -2/5 and then doing some manipulation

Dear Ant,
going from your first line to your second line you multiply. The equivalence between those two lines is lost if you multiply both sides by zero. In other words, the second line holds only if x is NOT -2/5 or 3/2 !

 

[SammyS]

Homework Statement

The question is stated at the top of the attached picture with a solution
20160303_095831.jpg
The correct results of the coefficients are A=2, B=-5, C=1
I have tried this problem multiple times and am still getting ugly coefficients. I have no idea why. A fresh pair of eyes would be welcome.
I have even had mathematica solve it for me to get some insight but it solves using matrices, which is not helpful when attempting it algebraically.

Your method looks perfectly fine to me.

 

[SteamKing]

Your method looks perfectly fine to me.

?

 

[AntSC]

From what i can see from textbooks the method I've chosen is valid. The reason behind making x=-2/5 and x=3/2 is to eliminate certain coefficients, leaving one unknown in each case thus making each coefficient solvable. Not quite sure of the logic of not dividing by those values, please explain if I'm missing something. Here's another question done the same way, which yielded the correct result - 20160303_134024.jpg. This agrees with the answers in the textbook and to double check, by solving with mathematica.
So my question is really why does this result produce correct answers and the question in my original post not?

 

[BvU]

Baffling that this works. Perhaps there is some logic in this and SteamKing and I should scratch behind our ears...

I did find the error in post #1: third line: $$x = -{2\over 5} \quad \Rightarrow \quad -{194\over 5} + 35 = -{19\over 5} C \quad \Rightarrow \ \ C =1 $$

 

[HallsofIvy]

The logic is that if 97x + 35 = A(5x + 2)^2 + B(2x-3)(5x+2) + C(2x-3) is true for all x then, in particular, it is true for x= -2/5 and x= 3/2.
Taking x= -2/5, 5x+ 2= 0 so the equation becomes -194/5+ 35= A(0)^2+ B(0)+C(-4/5- 3). Solve that for C.
Taking x= 3/2, 2x- 3= 0 so the equation become 291/2+ 35= A(15/2+ 2)^2+ B(0)+ C(0). Solve that for A.

To find B take x equal to any simple number, say x= 0.

 

[AntSC]

Not sure whether that's a sarcastic remark or not.
Thanks for spotting my lame mistake. Must have repeated it each time.
It all works out now. I got all 3 coefficients.
Perhaps we could talk further about the logic behind setting these x-values.
I understand your point of not allowing x=-2/5 and x=3/2, as in its original form would give undefined results. But once rearranged these x-values no longer have the same effect. How to resolve this?

 

[BvU]

No sarcasm. It's just that I agreed with SteamKing about the approach. But what you do appears to be equivalent.

Basically you want to solve $$97x + 35 = A(5x + 2)^2 + B(2x-3)(5x+2) + C(2x-3)\quad \forall x $$ for A, B and C. Where I wanted to exclude $x = -{2\over 5}$ and $x = {3\over 2}$.

Writing it all out gives three equations such as $25x^2 A + 10 x^2 B = 0$ and excluding those two specific x values doesn't change anything to the outcome.

And I read over the minus sign several dozen times too !

 

[Ray Vickson]

No sarcasm. It's just that I agreed with SteamKing about the approach. But what you do appears to be equivalent.

Basically you want to solve $$97x + 35 = A(5x + 2)^2 + B(2x-3)(5x+2) + C(2x-3)\quad \forall x $$ for A, B and C. Where I wanted to exclude $x = -{2\over 5}$ and $x = {3\over 2}$.

Writing it all out gives three equations such as $25x^2 A + 10 x^2 B = 0$ and excluding those two specific x values doesn't change anything to the outcome.

And I read over the minus sign several dozen times too !

If $97x + 35 = A(5x + 2)^2 + B(2x-3)(5x+2) + C(2x-3)$ for more than three distinct values of $x$, it holds for all $x$ (because a quadratic that is zero at 4 or more points is zero identically). So, it DOES hold at $x = -2/5$ and at $x = 3/2$. However, in this case the quadratic factor $(5x+2)^2$ means that we can differentiate both sides and evaluate again at $x = -2/5$ to obtain another relationship, giving 3 equations in the three unknowns $A,B,C$.

I don't look at posted handwritten work, so I don't know if the OP did that.

 

[BvU]

Hello Ray,

Since Steamking stated that $97x + 35 = A(5x + 2)^2 + B(2x-3)(5x+2) + C(2x-3)$ is correct, it looks as if I am the only one with this reservation about excluding $x = -{2\over 5}$ and $x = {3\over 2}$. Must be some result of being brainwashed to never ever multiply both sides of an equation by zero. And it indeed turns out to be completely harmless in this case.

I don't look at posted handwritten work, ...

I really like that. I'm not that far yet, but I do notice a growing preference for typed-out posts . Somewhere in the PF guidelines it's preferred, I know.
In defence of Ant: his(/her?) work is very neat and legible and the picture taking immaculate. I've seen far, far worse.

 

[Mark44]

Must be some result of being brainwashed to never ever multiply both sides of an equation by zero.

If you're going to be brainwashed, this is a good a concept as any.

The situation here is that the original equation is not equivalent to the one you get when all the denominators are cleared. In the original equation, x can't be 3/2, and it can't be -2/5. The revised equation has no restrictions at all. As HallsOfIvy said, the new equation has to be true for all values of x and for certain values of the parameters A, B, and C.

Regarding work posted as an image, there's a sticky at the top of this section saying that for a number of reasons, members should not post images of their work. For one thing, it's impossible for someone responding to insert a comment at the point where the work is wrong. For another, images are often posted sideways or upside down, are impossible or difficult to read, or are so messy with doodles and crossouts that they're hard to read.

 

[Ray Vickson]

Hello Ray,

Since Steamking stated that $97x + 35 = A(5x + 2)^2 + B(2x-3)(5x+2) + C(2x-3)$ is correct, it looks as if I am the only one with this reservation about excluding $x = -{2\over 5}$ and $x = {3\over 2}$. Must be some result of being brainwashed to never ever multiply both sides of an equation by zero. And it indeed turns out to be completely harmless in this case.

I really like that. I'm not that far yet, but I do notice a growing preference for typed-out posts . Somewhere in the PF guidelines it's preferrred, I know.
In defence of Ant: his(/her?) work is very neat and legible and the picture taking immaculate. I've seen far, far worse.

Re: about $x = -2/5$ and $x = 3/2$:
The quadratic $A(5x + 2)^2 + B(2x-3)(5x+2) + C(2x-3)$ is the numerator after putting all terms over a common denominator, so the equation $97x + 35 = A(5x + 2)^2 + B(2x-3)(5x+2) + C(2x-3)$ is required to hold at all $x$ other than -2/5 and 3/2 (because those two values of $x$ are excluded from the function's domain). However, as I said already, that implies that it hold for all $x$, including the "forbidden" values -2/5 and 3/2. It just so happens that those two values are particularly convenient to use.

Re: typed out work. I think you may have it backwards. It used to be that hardly anyone posted images (except for diagrams, etc.,) but recently it seems that more and more posters are trying to get away from typing things out. The pinned post "Guidelines for Students and Helpers" goes into this issue, and explains why posted images should be avoided as much as possible.

 

[SammyS]

When I first saw this method I responded much like SteamKing and BvU. How can you plug in values not in the domain of the original expressions?

I like to think of the rationale for the validity of this method as follows.

Starting with:

$\displaystyle \ \frac{97x+35}{(2x-3)(5x+2)^2}=\frac{A}{2x-3 }+\frac{B}{5x+2}+\frac{C}{(5x+2)^2} \$​

Multiply both sides by $\ (2x-3)(5x+2)^2\,, \$ giving:

$\displaystyle \ \frac{(97x+35)(2x-3)(5x+2)^2}{(2x-3)(5x+2)^2}=\frac{A(2x-3)(5x+2)^2}{2x-3 }+\frac{B(2x-3)(5x+2)^2}{5x+2}+\frac{C(2x-3)(5x+2)^2}{(5x+2)^2} \$​

Prior to cancelling common factors, both sides of this equation have the same implied domain as in the original equation: $\ x\ne\frac32 ,\ -\frac25$ .

However, for the expressions on either side of this equation, the discontinuities are removable. With suitable definitions to fill in those values, one can extend the functions on either side making each continuous on ℝ, and giving the result achieved by cancelling common factors.

$\displaystyle \ 97x+35=A(5x+2)^2+B(2x-3)(5x+2)+C(2x-3) \$​

.