What trick would you use in solving x^{2} -x^{3} = 12?
Cardano's method. I'm just not that tricky.What trick would you use in solving x^{2} -x^{3} = 12?
I'd use the rational root theorem. That's very straightforward, but "tricks" don't have to be otherwise.What trick would you use in solving x^{2} -x^{3} = 12?
Can you show how you'd factor it.[imath]x^2-x^3-12[/imath] is factorable or are you looking for a clever "trick"?
I believe the method of observation is valid and that gives you x = -2What trick would you use in solving x^{2} -x^{3} = 12?
Is -2 a triple root? If not, then you missed 2 roots.I believe the method of observation is valid and that gives you x = -2
If you plot f(x) = -x^3 + x^2 - 12, and put on your new glasses - you'll see that f(x) has only one real root.Is -2 a triple root? If not, then you missed 2 roots.
But but but... I like tedious Cardano!If you plot f(x) = -x^3 + x^2 - 12, and put on your new glasses - you'll see that f(x) has only one real root.
Since we found one real root, we can invoke polynomial division (\(\displaystyle \frac{-x^3+x^2-12}{x+2}\)) and get a quadratic quotient that will show the nature of other two roots (no need to invoke tedious Cardano)
That last step is not immediately obvious to most students, though it isn't hard once you think about it. You end up with -(x+2)(x^2-3x+6), if I did that right, and you do get the complex roots as well. (The same is true of the rational root approach.)I guess that the rational root theorem is the best way to solve this.
I watch a video to by a Nigerian mathematician who solved it very differentially. For the record, the only reason why I mentioned that this professor is from Nigeria is because he was taught differently and hence thinks differently those people from other areas.
He said to think of x^{2}-x^{3}= 12 as x^{2} - x^{3} - 2^{2} - 2^{3}=0 and factor the lhs.
I guess this was overkill.
The two missed roots I was referring to were complex roots.How about complex roots?
Possibly, in Nigeria the average student would think this way. I wonder. I enjoyed watching a number of youtube videos from this person, mathsonline tv, but he does do things differently--sometimes very differently.That last step is not immediately obvious to most students, though it isn't hard once you think about it.
I was referring to the factoring step, which is peculiar because two terms are constants; it's the usual factoring by grouping, but with a twist, namely that one pair factor as a sum of cubes rather than just a GCF. (Or is there another method?)Possibly, in Nigeria the average student would think this way. I wonder. I enjoyed watching a number of youtube videos from this person, mathsonline tv, but he does do things differently--sometimes very differently.
This is just the standard "ac grouping" method that applies to any quadratic, applied to a case where its full power is not needed. I've told students that if they want to memorize one method, for all quadratics, this is fine, though often you'll say at the end, "I coulda just written (x+2)(x+3)!" Not bizarre, but overkill.He factored in a very unusual way--by always grouping.
For example, in factoring x^{2}+ 5x + 6 he would say that you need to find two numbers that multiply out to 6 and add up to 5. The numbers he would say are 2 and 3. Then he would say x^{2}+ 5x + 6 = (x^{2}+ 2x) + (3x+ 6) and factor that--very bizarre in my opinion. In the end he gets the correct factorization.
[imath]\left\{ \begin{gathered} {x^2} - {x^3} = 12 \\ - {x^3} + {x^2} - 12 = 0 \\ {x^3} - {x^2} + 12 = 0 \\ \end{gathered} \right.[/imath] By inspection of the last, we see that [imath]x=-2[/imath] is a root.What trick would you use in solving x^{2} -x^{3} = 12?
He grouped it as (x^{2}-2^{2}) - (x^{3} + 2^{3})-----so he factored using the sum of cubes and the difference of squares.This is just the standard "ac grouping" method that applies to any quadratic, applied to a case where its full power is not needed. I've told students that if they want to memorize one method, for all quadratics, this is fine, though often you'll say at the end, "I coulda just written (x+2)(x+3)!" Not bizarre, but overkill.
Did he factor x^{2} - x^{3} - 2^{2} - 2^{3} using a sum of cubes as I mentioned, or another way?