4
$\begingroup$

I found this question on my test:

What is the sum of all of the real root of $x^3-4x^2+x=-6$?

  • A.) $-4 $
  • A.) $-2$
  • A.) $-0 $
  • A.) $2 $
  • A.) $4 $

My answer: $2 + 3 = 5$ , but that's not an option. Was the question wrong, or I didn't pay attention enough?

$\endgroup$
  • $\begingroup$ try to find the roots, they are all divisors of $6$ $\endgroup$ – Dr. Sonnhard Graubner Jan 15 '17 at 8:59
  • $\begingroup$ hint: the sum is $4$ $\endgroup$ – Dr. Sonnhard Graubner Jan 15 '17 at 9:00
  • $\begingroup$ Hint: There is a small negative root you haven't discovered. $\endgroup$ – Improve Jan 15 '17 at 9:00
  • 18
    $\begingroup$ @ Dr. Sonnhard Graubner: Well, I don't really think that's a hint. $\endgroup$ – Adola Jan 15 '17 at 9:05
6
$\begingroup$

$$x^3-4x^2+x=-6 \implies x^3-4x^2+x+6=0 \implies (x-2)(x-3)(x+1)=0 $$

You missed one solution, $x=-1$. Thus, the answer is $4$.

Some people may suggest that you use Vieta's formula, but IMO that would be unwise.

This is because Vieta's formula adds all the solutions, even the complex ones, but the question at hand explicitly asks for only real solutions.

So this would probably be the best way to do it.

$\endgroup$
  • $\begingroup$ @S It doesn't matter at all that Vieta's formulas adds "all the roots": if the polynomial is a real one, at the end all the Vieta's formulas, including the one with the sum of all the roots, with render real numbers, of course. $\endgroup$ – DonAntonio Jan 15 '17 at 9:07
  • 6
    $\begingroup$ @DonAntonio The sum would be a real number, but the roots may not be a real number, which is what I am trying to say. The questions asks for the sum of the real roots. $\endgroup$ – S.C.B. Jan 15 '17 at 9:08
  • $\begingroup$ @S Good catch, I didn't even notice that in the question. Thanks. +1 $\endgroup$ – DonAntonio Jan 15 '17 at 9:09
  • $\begingroup$ You are partly right about Vieta's formula. However, once you know that all but possibly one roots are real (and OP did find two real roots), it follows that they are all real (because nonreal roots would come in conjugate pairs), and you can apply Vieta's formula (which is much easier in general than looking for the last root). $\endgroup$ – tomasz Jan 15 '17 at 15:07
  • $\begingroup$ @tomasz Yes, you can. I am saying that you shouldn't apply Vieta's formula right away. And when it is a cubic, I think finding the last root and using Vieta's formula is really of similar difficulty. $\endgroup$ – S.C.B. Jan 15 '17 at 15:11
2
$\begingroup$

It is easy to find these roots: $-1,2,3$, so the sum is $4$.

$\endgroup$

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.