Polynomial-related question (why does this work algebraically?)

[lockem]

Alright, so this problem has been bothering me for a few days. I've asked three friends for their input, and they're just as stumped as me. I came across this problem in a "college algebra" practice book. What I mean by "this problem has been bothering me" isn't that I'm stuck/haven't found the solution; I've found the solution, it just doesn't seem to work, algebraically. After "solving" the problem, I referred to the solution in the book, and the author did the exact same thing as me. The issue I'm having, like I said, is that the solution to this problem (apparently, as far as I know) does not work algebraically.

Given the following equation, evaluate the expression a + 3b + 2c

2(x^2 - 4x + a)-3(-2x^2 + bx - 1)+5(cx^2 + 5x + 6)=23x^2 + 17x - 5

As far as I know, if you perform an operation on a term on the opposite side of an equation, you must perform that same operation on every term on the opposite side of the equation; to keep the equation "balanced." The (apparent) solution to this problem requires you to basically selectively/sequentially divide terms on the right side, by the corresponding term on the left side to find values for a, b and c. So, algebraically, the solution to this (again, as far as I know) does not seem to work. I'm hoping someone can enlighten me here.

2(x^2 - 4x + a)-3(-2x^2 + bx - 1)+5(cx^2 + 5x + 6)=23x^2 + 17x - 5

After distributing 2, -3, and 5 to the appropriate quantities, combining like-terms and reorganizing/grouping the terms according to the order they appear on the right side, I get:

(5cx^2 + 8x^2)+(-3bx + 17x)+(2a + 33)=23x^2 + 17x - 5

The next step I take is to combine the non-abc terms on the left, with the corresponding terms on the right. So I end up with:

5cx^2 - 3bx + 2a=15x^2 + 0x - 38

(I realize you don't normally write 0x, I just chose to do so in an attempt to illustrate more clearly what I'm trying to do here)

Next, keeping the ordering of the right side of the equation in mind, I divided 5cx^2 by itself to isolate c, and then divided 15x^2 by 5x^2 to get 3. Then -3bx by itself to isolate b, and obviously anything divided by 0=0, so moving onto a. I divided 2a by itself to isolate a and then -38/2 = -19. So:

a=-19
b=0
c=3

Plugging the numbers into the original "evaluate the expression a + 3b + 2c" I end up with a + 3b + 2c = -13. Basically, I just want to know why this solution works, because it seems to contradict my current understanding of algebra. Any help is greatly appreciated! Thanks.

 

[Ray Vickson]

I can't follow your reasoning, bit it seems needlessly complicated. First, though, you need to distinguish between an *equation* and an *identity*. Your equation has two roots x for ANY values of a, b and c. However, if you have, instead, an identity, it is supposed to be true for ALL values of x. So, if you have an identity of the form [tex] Ax^2 + Bx + C \equiv Dx^2 + Ex + F [/tex] then the coefficients must be equal. So we need A = D, B = E and C = F.

RGV

 

[Underhill]

Next, keeping the ordering of the right side of the equation in mind, I divided 5cx^2 by itself to isolate c, and then divided 15x^2 by 5x^2 to get 3. Then -3bx by itself to isolate b, and obviously anything divided by 0=0, so moving onto a. I divided 2a by itself to isolate a and then -38/2 = -19. So:

a=-19
b=0
c=3

Lockem,

I followed your reasoning until this paragraph. We have the 4-variable equation:

5cx[itex]^{2}[/itex]-3bx+2a = 15x[itex]^{2}[/itex]-38

It is impossible to solve this equation for anyone variable except in terms of the other three variables. How did you proceed? Can you detail your steps?

Underhill

 

[Ray Vickson]

Lockem,

I followed your reasoning until this paragraph. We have the 4-variable equation:

5cx[itex]^{2}[/itex]-3bx+2a = 15x[itex]^{2}[/itex]-38

It is impossible to solve this equation for anyone variable except in terms of the other three variables. How did you proceed? Can you detail your steps?

Underhill

I think one problem is that you are mixing up the concepts of variables and parameters (although, I admit, you are *allowed* to do that; it is just not a good idea). In the normal algebraic language, your equation has only ONE *variable* in it---namely, x---but also has three *parameters* a, b and c. Even if you also regard a, b and c as variables, there is no problem: we can solve for anyone in terms of the other three (as you said).

However: the real issue (which you have failed to address) is whether or not it is an equation or an identity. If it is an equation for x, it is just a quadratic equation of the form A*x^2 + B*x + C = 0; I'll let you figure out what A, B and C must be. Anyway, being a quadratic, it has two roots x = x1 or x = x2; here they are:
x1 = [3b+r]/[10(c-3)] and x2 = [3b-r]/[10(c-3)], where
r = sqrt(9b^2 + 2280 + 120a - 760c -40ac). It does not matter what a, c and c are: the roots x1 and x2 are computable in terms of them by an explicit formula. Of course, for some a, b and c combinations we may get *complex* roots rather than real ones, but they are still roots in the algebraic sense.

On the other hand, if your equation is supposed to be an _identity_ that holds for all x, then we need: 5c = 15, so c = 3, -3b = 0, so b = 0, and 2a = -38, so a = -19. Obviously, it cannot be an _identity_ in all of x, a, b and c, because it cannot hold for all values of a, b or c---only for the ones given above.

RGV

 

[DivisionByZro]

Next, keeping the ordering of the right side of the equation in mind, I divided 5cx^2 by itself to isolate c, and then divided 15x^2 by 5x^2 to get 3. Then -3bx by itself to isolate b, and obviously anything divided by 0=0.

1 - What does "... divided by itself..." mean? You can't just pick a term and divide it by whatever you want.

2 - "anything divided by zero" is NOT zero. 0 divided by something is zero, not the other way around. Also as the above poster mentioned, you have one equation in four unknowns. You won't be able to solve this for any particular values directly. If you're lucky you can randomly pick some numbers and take a guess, but that's about it.

I also do not follow all of your reasoning; basically your "trick" doesn't even work here. If you did get the right answer, then you're darn lucky. You methods are incorrect.