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Quadradic
For quadradic roots a, b if can be factored in (x-a)(x-b) and expanding is (x²-xb)-(ax-ab)
(x²-bx)+(ab-ax)
(((x²)+(ab))-(bx))-ax
(x²)+(ab)-(b+a)x
x²-(b+a)x+ab
However quadradic term is1
For ax²+bx+c=0
Idea
x²+(b/a)x+(c/a)=0
x_1+x_2=b/a
x_1x_2=c/a
To make it easier
Subsitute d=b/a, e=c/a (not euler's number)
x_1+x_2=d
x_1x_2=e
(d/2-j)+(d/2+j)=d
x_1x_2=e
(d/2-j)(d/2+j)=e
(d²/4-j²)=e
(-j²)=e - d²/4
(j²)=d²/4 - e
(j)=sqrt(d²/4 - e)
(d/2 - sqrt(d²/4 - e))(d/2 + sqrt(d²/4 - e))=d
(d/2 - sqrt(d²/4 - e)), (d/2 + sqrt(d²/4 - e)) is (d/2 ± sqrt(d²/4 - e))
(b/2a ± sqrt(b²/4a² - c/a))
(b/2a ± sqrt(b²/4a² - 4ca/4a²))
Quadradic formula usually has 4ac
(b/2a ± sqrt(b²/4a² - 4ac/4a²))
(b/2a ± sqrt((b² - 4ac)/4a²))
(b/2a ± sqrt(b² - 4ac)/sqrt(4a²))
(b/2a ± sqrt(b² - 4ac)/sqrt(4a²))
d=-b/a it has - in linear terms
(-b/2a ± sqrt(b²/4a² - c/a))
(-b/2a ± sqrt(b²/4a² - 4ac/4a²))
(-b/2a ± sqrt((b² - 4ac)/4a²))
(-b/2a ± (sqrt(b² - 4ac)/2a))
(-b ± (sqrt(b² - 4ac)))/2a
Sum of roots: b/a (from old d)
Product of roots: c/a
For b²-4ac<0: both are complex
For b²-4ac=0: 1 solution
For b²-4ac>0: 2 real solution

3 months ago | [YT] | 0