Is this way of teaching the quadratic solutions really anything new?

[swampwiz]

https://getpocket.com/explore/item/mathematician-finds-easier-way-to-solve-quadratic-equations

This seems to just be the quadratic formula in a transposed way.

 

[jedishrfu]

I think its a step beyond or between factoring in the traditional sense and using quadratic formula. It brings out a symmetry in the roots that isn't immediately obvious.

The is not unlike a similar symmetry found in complex roots of a polynomial being equally spaced about a circle in the complex plane aka DeMoivre’s theorem.

https://en.m.wikipedia.org/wiki/De_Moivre's_formula

i remember using the symmetry in a test because i couldn't remember the formula but did have one real root of the 5th order polynomial and we were to graph the solution.

 

[FactChecker]

It looks to me like a disguised process of completing the square, which is (IMO) an intuitive way to solve it or to develop the quadratic formula.

 

[jedishrfu]

The difference I see though is going to the sum and saying the roots are equidistant from half the sum.

When I learned to factor they didnt teach that instead you were looking for two numbers which summed correctly and when multiplied got the c term. It seems that root summing symmetry gets you to an answer more quickly.

 

[Mark44]

The difference I see though is going to the sum and saying the roots are equidistant from half the sum.

The technique described is not really all that different from what happens in the technique of completing the square. Let's look at an example, such as solving the equation $x^2 - 4x - 5 = 0$

If we approach this using completing the square, we have
$x^2 - 4x = 5$
$x^2 - 4x + 4 = 5 + 4$
$(x - 2)^2 = 9$
So either x - 2 = 3 or x - 2 = -3, producing the solutions x = 5 or x = -1.
The only difference I see is the substitution of $u$ for $x - 2$, so $u^2 = 9 \Rightarrow u = \pm 3$.

Undo the substitution, and you get the same solutions I showed.

 

[FactChecker]

The technique described is not really all that different from what happens in the technique of completing the square.

It seems like a geometric interpretation of the algebraic process of completing the square. I find the algebraic process to be direct enough that a geometric interpretation is only distracting. But people who are not as comfortable with the algebraic process might benefit from a geometric interpretation.

 

[uart]

I can see some benefits for early students in terms connecting it to the parabola axis of symmetry (-b/2a), and also some good practice with difference of two squares expansion/factorization. Also could be useful in introducing students to the concept of substitution of a variable in an equation.

I wouldn't want to fully replace teaching completing the square though, as that's also very useful for things other than solving quadratics. For example putting an integral like this one into a more amenable form.
\int \frac{1}{x^2 - 2x + 10} \, dx = \int \frac{1}{(x-1)^2 + 3^2} \,dx

BTW. I've used this technique many times in the past. Usually in the context of students first learning to factorize simple integer coeff quadratics in the form x^2 + bx + c with the old "sum = b and product = c" method.

After giving the usual simple exercises like x^2 + 8x + 15\, or x^2 + 5x + 6, better students will sometimes ask, "what if you just cannot find any two such numbers with the required sum and product?". The answer I give is that either no such numbers exist (no real roots) or that they exist but are surds and hence more difficult to find. I give an example like x^2 + 6x +7\, and tell them to try (-3+s) and (-3-s), where "s" is the surd to be determined. I never realized I'd found a "new process".

 

[symbolipoint]

I find the algebraic process to be direct enough that a geometric interpretation is only distracting. But people who are not as comfortable with the algebraic process might benefit from a geometric interpretation.

I found the simple geometric method to be extremely helpful, to ME at least, if not to other people.

 

[FactChecker]

I found the simple geometric method to be extremely helpful, to ME at least, if not to other people.

I'll buy that.