Timeline for Best way to explain the thinking steps from x² = 9 to x=±3
Current License: CC BY-SA 4.0
13 events
| when toggle format | what | by | license | comment | |
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| Sep 3, 2024 at 20:07 | comment | added | Yakk | @JiK So let's work over the Dual Numbers instead of the Reals (add z non-zero with z^2=0). We don't have ab=0 <=> to a=0 or b=0 now. Do we have x^2=9 imply x=-3 or +3? I'm thinking yes; so what we actually need for this trick to work isn't something I'm exactly sure of. But it isn't "free", you can have very real-number like things that don't have the property you want. | |
| Sep 3, 2024 at 18:49 | comment | added | Mahmoud | @JiK Oh, I see. I misunderstood your comment. | |
| Sep 3, 2024 at 14:53 | comment | added | JiK | @mhdadk If $p(x)$ has a factorization where $x-7$ is a factor, then $p(7)=0$. If $(x-3)(x+3)$ had another factorization that had $x-7$ as a factor, then $(x-3)(x+3)$ would be zero when $x=7$. | |
| Sep 3, 2024 at 14:10 | comment | added | Mahmoud | @JiK I believe you meant "...because then $(x-3)(x+3)$ wouldn't be zero when $x=7$..." and not "would". | |
| Sep 3, 2024 at 12:31 | comment | added | JiK | A smart student will realize that this means there can't be a factorization that has e.g. $x-7$ as a factor because then $(x-3)(x+3)$ would be zero when $x=7$ But this insight is not needed. | |
| Sep 3, 2024 at 12:28 | comment | added | JiK | @Yakk But you don't need to know about unique factorization for this. It's enough to know that $x^2-9$ and is zero if and only if one of the factors in this factorization is zero, because $ab=$ iff $a=0$ or $b=0$. If there happens to be another factorization, we still know that $(x-3)(x+3)$ can't be zero if $x$ is not $3$ or $-3$. | |
| Sep 2, 2024 at 17:06 | comment | added | Yakk | The fact you can uniquely factorize polynomials within the real numbers isn't "free"... and uniquely factoring to linear terms isn't even true! | |
| Sep 2, 2024 at 13:33 | comment | added | Wastrel | @mhdadk Well said. They can't be both, but they can be either. | |
| Sep 1, 2024 at 20:59 | comment | added | Mahmoud | @Wastrel I believe your second point can be proven more formally by contradiction: suppose that both $x-3 = 0$ and $x + 3 = 0$. This implies that $x = 3$ and $x = -3$. However, by the transitivity property of equality, this implies that $3 = -3$, which is a contradiction. Hence, we must have that either $x = 3$ or $x = -3$, but not both. | |
| Sep 1, 2024 at 14:14 | comment | added | Wastrel | I agree with the answer but I don't agree completely with the explanation. The hard part is teaching how to factor x^2−9. The student must learn to factor the difference of two squares. Once you have the two factors, one must be equal to zero for their product to be zero. Although 0*0=0, it can't be both, and that is easy to show, because it's simply "unfactoring" back to the original x^2-9. Then you can show that either factor can be equal to zero, and arrive at the two solutions. | |
| Sep 1, 2024 at 13:51 | comment | added | Steven Gubkin | This is also the closest to the formal proof using real number axioms, using trichotomy and existence of multiplicative inverses. | |
| Sep 1, 2024 at 0:42 | comment | added | Sue VanHattum♦ | I would definitely use that for 𝑥^2=𝑥. But it doesn't feel like the best solution in this case. (I'll keep thinking about it, though. Maybe I'll come around...) | |
| Sep 1, 2024 at 0:33 | history | answered | TomKern | CC BY-SA 4.0 |