Timeline for Why, in the rules regarding the discriminant determining whether roots are real or not, does it matter that the quadratic has real coefficients?
Current License: CC BY-SA 4.0
10 events
| when toggle format | what | by | license | comment | |
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| Dec 7, 2021 at 21:12 | vote | accept | CommunityBot | ||
| Dec 7, 2021 at 6:47 | answer | added | dxiv | timeline score: 3 | |
| Dec 7, 2021 at 6:09 | comment | added | dxiv | @poetasis Point was that your comment does not hold true in general, for arbitrary complex coefficients $a,b,c$. Mine was a counterexample to that effect, and a different one is given in Arturo's first comment. | |
| Dec 7, 2021 at 5:21 | comment | added | poetasis | @dxiv By your resoning, all equations have imaginary coefficients. $$ i(x^2 - 3x + 2)\quad = \quad ix^2 - 3ix + 2i = 0 $$ | |
| Dec 7, 2021 at 4:03 | comment | added | dxiv | @poetasis But for example $\,ix^2-3ix+2i=0\,$ has real roots (and negative discriminant). | |
| Dec 7, 2021 at 1:25 | comment | added | poetasis | Given $\quad x=\dfrac{-b\pm\sqrt{b^2-4ac}}{2a}\quad $ the $b$ outside the radical makes the roots imaginary in any case. | |
| Dec 6, 2021 at 22:37 | comment | added | user839943 | @dxiv I see. That explains why Arturo's solution is the better/correct one. Concerning "rules regarding the discriminant", I've only being introduced to the discriminant advising if the solution to a quadratic is real, double root or not real. | |
| Dec 6, 2021 at 22:20 | comment | added | dxiv | It is not clear which "rules regarding the discriminant" the question is asking about. If the coefficients are not real then the discriminant will not be real in general, so you cannot compare it to $0$, or say it is positive or negative. | |
| Dec 6, 2021 at 20:17 | comment | added | Arturo Magidin | If the coefficients are not real, and $b^2-4ac\gt 0$, you could still have no real roots because $b$ or $a$ are complex. For example, $x^2+ix-1$ has $b^2-4ac=-1+4=3\gt 0$, but it does not have real roots, as the roots are $\frac{\sqrt{3}-i}{2}$ and $\frac{-\sqrt{3}-i}{2}$. So even though the discriminant is positive, the roots are nonreal complex. | |
| Dec 6, 2021 at 20:15 | history | asked | user839943 | CC BY-SA 4.0 |