This seems more complicated and roundabout than completing the square. The average of the roots shows up that way too:
We want to write x^2 + bx + c = 0 in the form (x+m)^2 + n = 0, so there's only one x left and we can rearrange for it.
Expanding, (x+m)^2 = x^2 + 2mx + m^2, so we get the x^2 we want, and the coefficients of x tell us b = 2m, so m = b/2. We also get an m^2 (= b^2/4) we don't want, so let's take it away:
(x + b/2)^2 - b^2/4 = x^2 + bx
That x + b/2 is x - (-b/2), x minus the average of the roots, which is the x value the parabola is centred on. Then we add c:
(x + b/2)^2 - b^2/4 + c = x^2 + bx + c
To find the roots, set it to 0 and rearrange for the one x that's left:
(x + b/2)^2 - b^2/4 + c = 0
(x + b/2)^2 = b^2/4 - c
x + b/2 = ±√(b^2/4 - c)
x = -b/2 ± √(b^2/4 - c)
Note that this is the average of the roots ± the article's z. Then combine:
x = -b/2 ± √((b^2-4c)/4)
x = -b/2 ± √(b^2-4c)/2
x = (-b ± √(b^2-4c))/2
If you have ax^2+bx+c = 0, divide the equation by a first, then do the same steps and you get the normal quadratic formula:
x = (-b ± √(b^2-4ac))/(2a)
I think the linked post misstates the purpose of the article: it's not about new maths, but about pedagogy and ways of explaining the quadratic formula.
Having taught elementary algebra for many years at a community college I think your perception of what is complicated and easy are not correct. The method in the article is far easier for elementary algebra students than what you describe.
You’ve proved the quadratic formula. This is a formula students in elementary algebra will struggle with. It’s a formula whose proof will be lost on them. What is shown in the article is an easy to apply mechanism for finding the roots. The method in the article is one that I can use in the classroom. There’s no way I’d attempt your explanation in an elementary algebra class.
I wouldn't try to teach it to elementary algebra students either; I was explaining for maths-weak HN readers. It would be worth trying with some of my first year university students (but definitely not all). What's shown in the article is also a derivation of the quadratic formula. I'm not convinced the less general solving method at the end of the article is any easier than completing the square, and it's certainly more mystifying, as completing the square can be checked by expanding, but this is a bit out of nowhere. I will try the article's method if I get an opportunity, but I think the introduction of an extra variable z is quite an obstacle, and that the fact that -b/2 is the average of the roots will be quite a leap for most students.
Comments
This seems more complicated and roundabout than completing the square. The average of the roots shows up that way too:
We want to write x^2 + bx + c = 0 in the form (x+m)^2 + n = 0, so there's only one x left and we can rearrange for it.
Expanding, (x+m)^2 = x^2 + 2mx + m^2, so we get the x^2 we want, and the coefficients of x tell us b = 2m, so m = b/2. We also get an m^2 (= b^2/4) we don't want, so let's take it away:
(x + b/2)^2 - b^2/4 = x^2 + bx
That x + b/2 is x - (-b/2), x minus the average of the roots, which is the x value the parabola is centred on. Then we add c:
(x + b/2)^2 - b^2/4 + c = x^2 + bx + c
To find the roots, set it to 0 and rearrange for the one x that's left:
(x + b/2)^2 - b^2/4 + c = 0
(x + b/2)^2 = b^2/4 - c
x + b/2 = ±√(b^2/4 - c)
x = -b/2 ± √(b^2/4 - c)
Note that this is the average of the roots ± the article's z. Then combine:
x = -b/2 ± √((b^2-4c)/4)
x = -b/2 ± √(b^2-4c)/2
x = (-b ± √(b^2-4c))/2
If you have ax^2+bx+c = 0, divide the equation by a first, then do the same steps and you get the normal quadratic formula:
x = (-b ± √(b^2-4ac))/(2a)
I think the linked post misstates the purpose of the article: it's not about new maths, but about pedagogy and ways of explaining the quadratic formula.
Having taught elementary algebra for many years at a community college I think your perception of what is complicated and easy are not correct. The method in the article is far easier for elementary algebra students than what you describe.
You’ve proved the quadratic formula. This is a formula students in elementary algebra will struggle with. It’s a formula whose proof will be lost on them. What is shown in the article is an easy to apply mechanism for finding the roots. The method in the article is one that I can use in the classroom. There’s no way I’d attempt your explanation in an elementary algebra class.
I wouldn't try to teach it to elementary algebra students either; I was explaining for maths-weak HN readers. It would be worth trying with some of my first year university students (but definitely not all). What's shown in the article is also a derivation of the quadratic formula. I'm not convinced the less general solving method at the end of the article is any easier than completing the square, and it's certainly more mystifying, as completing the square can be checked by expanding, but this is a bit out of nowhere. I will try the article's method if I get an opportunity, but I think the introduction of an extra variable z is quite an obstacle, and that the fact that -b/2 is the average of the roots will be quite a leap for most students.