Prove that if ax^2 + bx + c = dx^2 + ex + f, then...?

Question

This is especially required when doing partial fractions in Calculus. However, it was something I just accepted without ever seeing a proof that it's true.

How do you prove that, for real numbers a,b,c,d,e,f, that if
ax^2 + bx + c = dx^2 + ex + f, then
a = d, b = e, and c = f?

I tried doing it last night and couldn't get anywhere. I would love to see a rigorous proof of this from one of the many math geniuses on here.

Answer 1

Suppose that ax^2 + bx + c = dx^2 + ex + f for all real numbers x.

If we bring everything to one side, we get

ax^2 + bx + c - dx^2 - ex - f = 0

or

(a - d)x^2 + (b - e)x + (c - f) = 0.

Remember that this holds for all real x.

Thus, every real number x is a root of the polynomial (a - d)x^2 + (b - e)x + (c - f).

But any polynomial of degree 2 has at most two roots; any polynomial of degree 1 has at most one root; and any nonzero polynomial of degree 0 has no roots.

Thus, it must be that (a - d)x^2 + (b - e)x + (c - f) is the zero polynomial. Therefore, a - d = 0, b - e = 0, and c - f = 0. Therefore, a = d, b = e, and c = f.

Answer 2

I agree with the former answers, if you want the both expressions to be equals forall x, then there is a theorem that proves that. But if you only want to find some value(s) so that ax^2 + bx +c is equals to dx^2 + bx + c, this is not true.

Consider this example:

x^2 + 2x + 1 is 0 if x= -1

And 2x^2 + 4x + 2 is 0 too.

So, you have two different expresions and their value is the same for a due x.

Another example could be this one:

(x-4)(x+8) = 50(x-4)(x+1), if x = 4, both sides are equals to 0

Ana

Answer 3

ax^2 + bx + c = dx^2 + ex + f
ax^2 - dx^2 + bx - ex + c - f = 0
x^2(a - d) + x(b - e) + (c - f) = 0

As you see, the only way to make every part of this problem equal to zero is if a = d , b = e , and c = f. Otherwise, it would result in a non-zero answer. Remember, if c = f, then x could feasibly be zero, so be careful.

Answer 4

You are missing a slight detail that has major repurcussions. YOu want ax^2+bx+c to be EQUIVALENT to dx^2+ex+f. In other words, you want them to be equal for ALL x. Now the proof is simple. I will start you off with the linear case, which easily generalizes to all polynomials, and is called "equating coefficients".

Let ax+b = cx+d for all x. Suppose b did not equal d. Then a(0)+b does not equal c(0)+d, contradiction. So b=d. Then we can also state that ax=cx for all x. This is then true for x=1, so a=c. Note that in generalizing, it might be helpful to perform induction on the degree of the polynomials in x.

Steve

Answer 5

omg! that's crazy!! i don't know....
i can't do a simple calculus problem >-<
could you help me? i asked a question a few minutes ago. pleeeseeeeeeeeeeeeeeee?