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Linear
ax+b=0
ax=-b
x=-b/a
Quadradic
Ax²+bx+c
x²+(b/a)x+(c/a)=0
Substitute x=y-h
A(y-h)²+b(y-h)+c what h make linear coefficient 0 is b/2a
So we Substitute x=y-(b/2a)
This looks like something
(y-(b/2a))²+(b/a)(y-(b/2a))+(c/a)
y²+(4ac-b²)/(4a²)=0
y²=-(4ac-b²)/(4a²)
y=±sqrt(-(4ac-b²)/(4a²))
y=±sqrt(-(4ac-b²))/2a
y=sqrt((b²-4ac))/2a
x=y-(b/2a)->x+(b/2a)=y
x+b/2a=±sqrt((b²-4ac))/2a
x=(-b±sqrt((b²-4ac)))/2a
Cubics
ax³+bx²+cx+d
x³+(b/a)x²+(c/a)x+(d/a)=0
A(x-h)³+b(x-h)²+c(x-h)+d what h make quadradic coefficient 0
b/3a
Plug h=b/3a is x³+px+q (p=(3ac-b²)/3a² and q=(2b³ – 9abc +27a²d)/27a³)
x³+px+q=0
x=u+v
(u+v)³+p(u+v)+q=0
(u+v)³=u³+3u²v+3uv²+v³
(u+v)³=u³+3u²*v+3uv*v+v³
(u+v)³=u³+v(3u²+3uv)+v³
(u+v)³=u³+3v(u²+uv)+v³
(u+v)³=u³+3uv(u+v)+v³
u³+3uv(u+v)+v³+p(u+v)+q=0
u³+v³+3uv(u+v)+p(u+v)+q=0
u³+v³+(3uv+p)(u+v)+q=0
uv=-p/3
u³+v³+q=0
u³+v³=-q
uv=-p/3 and u³+v³=-q
u³v³=-p³/27 and u³+v³=-q
U=u³, V=v³
UV=-p³/27 and U+V=-q
t²+qt-p³/27
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3 months ago (edited) | [YT] | 0