# Proving that $ax^2 + bx + c = dx^2 + ex + f$

So given $ax^2 + bx + c = dx^2 + ex + f$ and that it holds true for all values of x:

Prove $a = d$, $b = e$, and $c = f$.

What I have done so far is set the equation equal to zero and factor the desired variables together and out:

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

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

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

Now once I prove that $a - d$, $b - e$, and $c - f$ are all equal to zero, then everything would be all over.

But unfortunately I do not know how to proceed here. I know it probably has to do with polynomial degree root limits or something, but I am fuzzy on the precise procedure and would appreciate the help on finishing this off.

• Do you know how to calculate the roots of a second order equation? If yes, do you know how many solutions you have to expect? – Thomas Sep 10 '16 at 10:51
• if this equation holds for all real $x$, so we can set $x=0$ and we get $$c=f$$ – Dr. Sonnhard Graubner Sep 10 '16 at 10:56

.Since the equation: $$(a-d)x^2 + (b-e)x+(c-f)=0$$ is true for all values of $$x$$, we can substitute values of $$x$$ and the resulting equation will still be true:

Put $$x=0$$, then we get $$c-f=0$$, so $$c=f$$.

Now our equation simplifies to : $$(a-d)x^2 + (b-e)x=0 \implies x((a-d)x+(b-e)) = 0$$

Take $$x=1$$ here: $$(a-d) + (b-e) = 0$$

Take $$x=-1$$ here:$$(d-a) + (b-e) = 0$$

Add the two equations: $$2(b-e) = 0$$, so $$b-e=0$$ and $$b=e$$

Finally, we are left with $$(a-d)x^2=0$$ for all values of $$x$$. Put $$x=1$$ to get $$a-d=0$$ so $$a=d$$. This method required no kind of complicated method and can be generalized to other polynomials.

• Thank you! However, I actually found a potential alternate method: using $(a-d)x^2 + x(b-e) = 0$ and dividing on both sides by x, then setting the $(a-d)x$ to zero, thus making $b - c = 0$. would this work as well? – KMoy Sep 10 '16 at 19:13
• Also, I believe that you wrote b = c?? Did you mean b = e? – KMoy Sep 10 '16 at 22:16
• Abosultely. This would work too. And thank you for pointing out the error. – астон вілла олоф мэллбэрг Sep 11 '16 at 3:24
• @KMoy I should add that division by $x$ should not be done because if $x=0$ then you can't divide. That's why I had to take two different values t that point, not just substitute $x=0$ and get a result. – астон вілла олоф мэллбэрг Sep 11 '16 at 7:33