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Math mistakes
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blabber
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nemY
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Kyuukyuu
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the six one is still pretty awesome though | ||
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conCentrate9
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AlwaysGG
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ninjafetus
United States231 Posts
Show why the vertex of f = ax^2 + bx + c must occur at x = -b / (2a) The idea was for them to remember that a vertex was a critical point, so they would take a derivative, set it equal to zero, and solve. Instead they wrote the quadratic formula x = (-b ± sqrt(b^2 - 4ac) ) / (2a) "expanded" as x = (-b + sqrt(b^2 - 4ac) - sqrt(b^2 - 4ac) ) / (2a) canceled the sqrt terms, and ended with x = - b / (2a). ..... .................. I almost gave them credit for such brilliant incompetence. | ||
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evanthebouncy!
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HeavOnEarth
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but noo. -pout- | ||
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Divinek
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MasterOfChaos
Germany2896 Posts
On May 18 2009 15:37 ninjafetus wrote: Instead they wrote the quadratic formula x = (-b ± sqrt(b^2 - 4ac) ) / (2a) "expanded" as x = (-b + sqrt(b^2 - 4ac) - sqrt(b^2 - 4ac) ) / (2a) canceled the sqrt terms, and ended with x = - b / (2a). I almost gave them credit for such brilliant incompetence. Why incompetence? It is a viable way to arrive at the solution if you recall that the vertex is in the middle of the arms. That is similar to how I arrived at that formula several years before learning what a derivative is. | ||
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klizzer
517 Posts
![[image loading]](http://roberdo.com/wp-content/uploads/2006/02/maths-problem.jpg) Also, the expanding thing above isn't necessarily wrong, it just lacks one step of counting the average, as MoC said. | ||
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ninjafetus
United States231 Posts
On May 18 2009 19:37 MasterOfChaos wrote: Why incompetence? It is a viable way to arrive at the solution if you recall that the vertex is in the middle of the arms. That is similar to how I arrived at that formula several years before learning what a derivative is. It's incompetence because A ± B = A + B - B = A doesn't make sense. You're implying that B =0, always. Yeah, there's other ways to get it (factor to vertex form), and, if we ignore the middle step, what they did will work in the specific case that the discriminant is zero, but other than that, it's just wrong. The reason I expected them to use the derivative is because 1) it was a calculus class, and 2) I had shown them the exact steps I was expecting before. If they did a legitimate factoring, I gave them full credit, of course. If they had averaged the two roots, I would have given them credit. That's not what they were thinking, I promise. | ||
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Musoeun
United States4324 Posts
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clazziquai
6685 Posts
They still do!  Thanks for sharing (again) ^_^ | ||
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