Prove that $x^3+px+q$ has multiple root $\Longleftrightarrow$ the discriminant is $0$. Is my solution correct?

Asked Apr 20 '14 at 09:07

Active Apr 20 '14 at 09:13

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$f'=x^2+p$

$f'=0,p=-3x^2$

Using $p=-3x^2,f=x^3-3x^3+q$

$f=0,q=2x^3$

Then the discriminant is: $$D=-108\cdot\big((2x^3/2)^2+(-3x^2/3)^3\big)=-108\cdot(x^6-x^6)=0$$

Even if it's correct, can you give me some explanation? I just don't know why I get this result.

edited Apr 20 '14 at 09:13

user2345215

asked Apr 20 '14 at 09:07

user128576

http://math.stackexchange.com/questions/432/why-does-the-discriminant-of-a-cubic-polynomial-being-less-than-0-indicate-compl – Prahlad Vaidyanathan Apr 20 '14 at 09:15
What is $f'$?. What do you think $p$ and $q$ are? – preferred_anon Apr 20 '14 at 09:21
Please rethink your computation of $D$. From sour ansatz, it seems you want to use $f$ has multiple roots iff $f$ and $f'$ have a common root? How is the discriminant define for you, just as a fancy expression in the coefficients or as something like $\prod(x_i-x_j)$? – Hagen von Eitzen Apr 20 '14 at 09:23
f' is the derivative of f,and i need to prove this by using the derivative not that product. – user128576 Apr 20 '14 at 09:26
I will share the whole task: – user128576 Apr 20 '14 at 09:29
Show with the derivative that x^3+px+q has multiple root exactly when the discriminant is 0. – user128576 Apr 20 '14 at 09:30
See cubic function. – Lucian Apr 20 '14 at 11:33

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