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.

  • $\begingroup$ 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? $\endgroup$ – Thomas Sep 10 '16 at 10:51
  • $\begingroup$ if this equation holds for all real $x$, so we can set $x=0$ and we get $$c=f$$ $\endgroup$ – 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.

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    $\begingroup$ 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? $\endgroup$ – KMoy Sep 10 '16 at 19:13
  • $\begingroup$ Also, I believe that you wrote b = c?? Did you mean b = e? $\endgroup$ – KMoy Sep 10 '16 at 22:16
  • $\begingroup$ Abosultely. This would work too. And thank you for pointing out the error. $\endgroup$ – астон вілла олоф мэллбэрг Sep 11 '16 at 3:24
  • $\begingroup$ @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. $\endgroup$ – астон вілла олоф мэллбэрг Sep 11 '16 at 7:33

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