- #1
protonchain
- 98
- 0
When I was much younger (back in like middle school or freshman year in HS) I always wondered how you prove the quadratic formula, and then a couple years later I just fiddled around and came up with:
[tex] Given: ax^2 + bx + c = 0 [/tex]
[tex] (\alpha x + \beta)^2 + \delta = 0 [/tex]
[tex] (\alpha x + \beta)^2 + \delta = \alpha^2 x^2 + 2 \alpha \beta x + \beta^2 + \delta = 0[/tex]
[tex] \alpha^2 x^2 + 2 \alpha \beta x + \beta^2 + \delta = ax^2 + bx + c [/tex]
[tex] \alpha^2 = a [/tex]
[tex] 2 \alpha \beta = b [/tex]
[tex] \beta^2 + \delta = c [/tex]
so
[tex] \alpha = \sqrt{a} [/tex]
[tex] \beta = \frac{b}{2 \sqrt{a}} [/tex]
[tex] \delta = c - \frac{b^2}{4a} [/tex]
[tex] (\alpha x + \beta)^2 + \delta = 0 [/tex]
[tex] (\alpha x + \beta)^2 = -\delta [/tex]
[tex] (\alpha x + \beta) = \pm \sqrt{-\delta} [/tex]
[tex] \alpha x = -\beta \pm \sqrt{-\delta} [/tex]
[tex] x = \frac{-\beta \pm \sqrt{-\delta}}{\alpha} [/tex]
Substituting gives
[tex] x = \frac{-\frac{b}{2 \sqrt{a}} \pm \sqrt{-(c - \frac{b^2}{4a}})}{\sqrt{a}} [/tex]
[tex] x = \frac{-\frac{b}{2 \sqrt{a}} \pm \sqrt{\frac{b^2}{4a} - \frac{4ac}{4a}}}{\sqrt{a}} [/tex]
[tex] x = \frac{-\frac{b}{2 \sqrt{a}} \pm \frac{\sqrt{b^2 - 4ac}}{2\sqrt{a}}}{\sqrt{a}} [/tex]
[tex] x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2 \sqrt{a} \sqrt{a}} [/tex]
[tex] x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2 a} [/tex]
I'm sure there's a much simpler way, I just never figured it out lol.
[tex] Given: ax^2 + bx + c = 0 [/tex]
[tex] (\alpha x + \beta)^2 + \delta = 0 [/tex]
[tex] (\alpha x + \beta)^2 + \delta = \alpha^2 x^2 + 2 \alpha \beta x + \beta^2 + \delta = 0[/tex]
[tex] \alpha^2 x^2 + 2 \alpha \beta x + \beta^2 + \delta = ax^2 + bx + c [/tex]
[tex] \alpha^2 = a [/tex]
[tex] 2 \alpha \beta = b [/tex]
[tex] \beta^2 + \delta = c [/tex]
so
[tex] \alpha = \sqrt{a} [/tex]
[tex] \beta = \frac{b}{2 \sqrt{a}} [/tex]
[tex] \delta = c - \frac{b^2}{4a} [/tex]
[tex] (\alpha x + \beta)^2 + \delta = 0 [/tex]
[tex] (\alpha x + \beta)^2 = -\delta [/tex]
[tex] (\alpha x + \beta) = \pm \sqrt{-\delta} [/tex]
[tex] \alpha x = -\beta \pm \sqrt{-\delta} [/tex]
[tex] x = \frac{-\beta \pm \sqrt{-\delta}}{\alpha} [/tex]
Substituting gives
[tex] x = \frac{-\frac{b}{2 \sqrt{a}} \pm \sqrt{-(c - \frac{b^2}{4a}})}{\sqrt{a}} [/tex]
[tex] x = \frac{-\frac{b}{2 \sqrt{a}} \pm \sqrt{\frac{b^2}{4a} - \frac{4ac}{4a}}}{\sqrt{a}} [/tex]
[tex] x = \frac{-\frac{b}{2 \sqrt{a}} \pm \frac{\sqrt{b^2 - 4ac}}{2\sqrt{a}}}{\sqrt{a}} [/tex]
[tex] x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2 \sqrt{a} \sqrt{a}} [/tex]
[tex] x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2 a} [/tex]
I'm sure there's a much simpler way, I just never figured it out lol.