Earlier quoted context omitted.
This is not about deep math. It's about pedagogics. How to generate some understanding and good feelings for solving quadratic equations.
I just don't get why making some unsupported postulate about how parabolas can be generated from straight lines is better pedagogically than teaching completing the square and providing a geometric interpretation for completing the square.
A new way to make quadratic equations easy (2019)
31–40 of 80 posts
Re: A new way to make quadratic equations easy (2019)
#32https://en.etudes.ru/models/conic-sections-cavalieri-parabol...
This may make all of the variables more "real" - but note that the equation graphed is really A(X - B)^2 = Y - C, where (B, C) is the vertex and A is the "setting" (clear if you examine the device. This is relatively similar to the complete-the-square formalism but with a straightforward geometric and "hands-on" interpretation.
Re: A new way to make quadratic equations easy (2019)
#33Earlier quoted context omitted.
I think, like the OP, this just dances around completing the square without actually eliminating the need for it. How do we know that all parabolas can be expressed as the product of two symmetrical straight lines?
This is not about deep math. It's about pedagogics. How to generate some understanding and good feelings for solving quadratic equations.
Re: A new way to make quadratic equations easy (2019)
#34I need to solve quadratic equations. It is much harder. Like
a1 x y + b1 x + c1 y + d1 = 0
a2 y z + b2 y + c2 z + d2 = 0
a3 z x + b3 z + c3 x + d3 = 0
Given a,b,c,d, solve it for x,y,z.
And in the general case where there are more than 3 equation or more than 3 x,y,z variables.
We can make a undirected graph with the variables as nodes and an edge if there is an equation containing those two variables. Like (x,y), (y,z), (x,z) here. If there is a cycle in that graph, we can combine the equations to obtain a single quadratic equation, which has two solutions for x.
I can search for cycles, and sometimes two cycles restrict each other, such that only one solution for x remains. Sometimes two solutions remain. I do not understand when that happens, and if there is a better way than searching cycles.
(also I can have an additional list of restrictions between the parameters. Like a1 = -c2. This sometimes leads to a cycle canceling itself out, revealing no information about x at all. )
Re: A new way to make quadratic equations easy (2019)
#35Earlier quoted context omitted.
This is not about deep math. It's about pedagogics. How to generate some understanding and good feelings for solving quadratic equations.
I honestly don’t see how this version is any more intuitive than regular completing the square.
Re: A new way to make quadratic equations easy (2019)
#36Re: A new way to make quadratic equations easy (2019)
#37Re: A new way to make quadratic equations easy (2019)
#38The author should really remove the words 'simple', 'easy', and 'intuitive' from this article. They clearly know nothing of why most people detest math. If you're already familiar with the quadratic equation and you care about math and even enjoy it you might find this new equation interesting, useful, or intuitive, but I guarantee it is still a garbled mass of un-intuitive numbers and symbols to "[m]any former algeb…
Re: A new way to make quadratic equations easy (2019)
#39For this very specific use case, I would call it sub-optimal, for the simple reason it takes all of five minutes to teach a class the formula by singing it to the tune of Frère Jacques. I haven't used the formula in over a decade but can still sing it.
Re: A new way to make quadratic equations easy (2019)
#40 x^2 + bx + c
to x^2 + bx + c + b^2/4 - b^2/4
This step is freaking weird, I've never seen it before, and I don't think this way. The process I follow is more like I observe that for general k: (x + k)^2 = x^2 + 2kx + k^2
so I observe that for k = b/2 I get (x + b/2)^2 = x^2 + bx + b^2/4
and then I figure out what I need to add to both sides of that equation to make the RHS equal the original quadratic, which gives me: (x + b/2)^2 - b^2/4 + c = x^2 + bx + c
and I then find the roots of the RHS by instead finding them for the LHS. We say that the LHS is the result of "completing the square" on the RHS.Question: Does anyone seriously learn to solve quadratics by rewriting x^2 + bx + c to x^2 + bx + c + b^2/4 - b^2/4. Because that seems like a bad way of learning it.