4 ms·

the beginning of the end of my math studies was eighth-grade algebra. i GOT algebra, it was easy, and i enjoyed it. ..until the quadratic equation was introdu

the beginning of the end of my math studies was eighth-grade algebra. i GOT algebra, it was easy, and i enjoyed it. ..until the quadratic equation was introduced. "hey, that's really neat! how does it work?" "oh, you'll learn that in calculus, which we won't allow you to take for another four years."

Chinjut 9y ago
It's common not to explain where the quadratic formula comes from (which is silly, it's straightforward enough to show, but standard math education curricula are shot through with this blind formula memorization nonsense), but… do they really tell you you'll learn it in calculus? It's got nothing to do with calculus.
jacobolus 9y ago
Not to mention, the quadratic formula per se is needlessly complicated. If you break it into two or three steps it makes a lot more sense: Step 1: move pieces around and divide by the leading coefficient to put equation into the form x² + b = 2ax (or if you like, x² – 2ax + b = 0; or feel free to swap the sign of b if you prefer). The equation for the parabola is then x² + b = 2ax + y. Step 2 (optional): rearrange that equation to get (x – a)² = a² – b Step 3: x = a ± √(a² – b) In this form, the “discriminant” is just a² – b, the x coordinate of the vertex is a and the y coordinate is b – a², Viète’s formulas tell us that the two roots satisfy ½(x₁ + x₂) = a and x₁x₂ = b. If the coefficients are real but the roots are complex, then we know each root has amplitude √b and phase arccos(a/√b). Etc.
- jacobolus 9y ago
  Or even better in some contexts is the form x² + b² = 2ax, in which case we have a = ½(x₁ + x₂) [the arithmetic mean of the roots] and b = √x₁x₂ [the geometric mean of the roots], and the discriminant is a² – b², which has a nice symmetry.

4 ms·

the beginning of the end of my math studies was eighth-grade algebra. i GOT algebra, it was easy, and i enjoyed it. ..until the quadratic equation was introdu

by steanne 9y ago

Chinjut 9y ago
It's common not to explain where the quadratic formula comes from (which is silly, it's straightforward enough to show, but standard math education curricula are shot through with this blind formula memorization nonsense), but… do they really tell you you'll learn it in calculus? It's got nothing to do with calculus.
jacobolus 9y ago
Not to mention, the quadratic formula per se is needlessly complicated. If you break it into two or three steps it makes a lot more sense: Step 1: move pieces around and divide by the leading coefficient to put equation into the form x² + b = 2ax (or if you like, x² – 2ax + b = 0; or feel free to swap the sign of b if you prefer). The equation for the parabola is then x² + b = 2ax + y. Step 2 (optional): rearrange that equation to get (x – a)² = a² – b Step 3: x = a ± √(a² – b) In this form, the “discriminant” is just a² – b, the x coordinate of the vertex is a and the y coordinate is b – a², Viète’s formulas tell us that the two roots satisfy ½(x₁ + x₂) = a and x₁x₂ = b. If the coefficients are real but the roots are complex, then we know each root has amplitude √b and phase arccos(a/√b). Etc.
- jacobolus 9y ago
  Or even better in some contexts is the form x² + b² = 2ax, in which case we have a = ½(x₁ + x₂) [the arithmetic mean of the roots] and b = √x₁x₂ [the geometric mean of the roots], and the discriminant is a² – b², which has a nice symmetry.