0
$\begingroup$

So I stumbled upon a question in a book that goes, "Prove that in any field, if $ax_1^2 + bx_1 + c = 0$ and $a ≠ 0$, then $x_2 = -(\frac {b}{a} + x_1)$ satisfies $ax_2^2 + bx_2 + c = 0$".

My question is, would merely substituting $x_2 = -(\frac{b}{a} + x_1)$ into the last equation be enough to complete the proof, or would I need to somehow derive the expression for $x_2$ from both the first and last equation?

$\endgroup$
  • 2
    $\begingroup$ Simple substitution is sufficient, although you'll probably get to a point where the first equation is necessary to prove you get zero $\endgroup$ – Michael Stachowsky Jun 24 at 0:57
  • $\begingroup$ @MichaelStachowsky Alright, that's what I did at first, though I wasn't sure if it was enough for a complete proof. Thanks! $\endgroup$ – Aden Jun 24 at 1:03
  • $\begingroup$ The statement "$x_2=-(\frac ba + x_1)$ satisfies $ax_2^2 + bx)2 + c= 0$" means exactly that plugging $-(\frac ba + x_1)$ into it gives you zero. And surely one way to proof that something happens is ... to do it and show it happens. If you want to prove an elephant can eat a penguin you can speculate all sorts of things about the metabolism and what have you, but if you actually filmed an elephant eating a penguin ... well... that would prove an elephant can eat a penguin, wouldn't it? Because it did.... $\endgroup$ – fleablood Jun 24 at 1:12
2
$\begingroup$

As per michael-stachowsky's comment, it suffices to substitute the value for $x_2$ into the equation to complete the proof, but it may be simpler to note that, if $a(x-x_1)(x-x_2)=ax^2+bx+c,$ then from comparing coefficients $-a(x_1+x_2)=b$.

$\endgroup$
  • $\begingroup$ I appreciate your answer, though if I may, is there another way to derive the expression for $x_2$ without comparing coefficients of a quadratic polynomial or using the quadratic formula? $\endgroup$ – Aden Jun 24 at 1:05
0
$\begingroup$

Alternatively: $$ax_1^2+bx_1+c=0=ax^2+bx+c \iff \\ a(x-x_1)(x-(-b/2-x_1))=0 \Rightarrow \\ x_1=x_1, x_2=-b/2-x_1.$$

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.