# Linear formula proof:
Given ax + b = 0 and a =/= 0, we have the following: x + b/a = 0, so x = - b/a.
# Quadratic formula proof:
Given ax² + bx + c = 0 and a =/= 0, we have the following:
x² + bx/a + c/a = 0, so x² + bx/a + (b/2a)² + c/a = (b/2a)²,
so x² + 2*x*(b/2a) + (b/2a)² = (bx/2a)² - c/a,
so (x + b/2a)² = (b/2a)² - c/a, so (x + b/2a)² = b²/4a² - 4ac/4a²,
so (x + b/2a)² = (b² - 4ac) / 4a², so x + b/2a = +- ²√(b² - 4ac) / ²√(4a²),
so x = -b / 2a +- √(b² - 4ac) / 2a = [ -b +- √(b² - 4ac) ] / 2a.
# Cubic formula proof:
*** See after the following quartic formula proof. ***
# Quartic formula proof:
Given ax⁴ + bx³ + cx² + dx + e = 0 = (Ax² + Bx + C)² - (Dx² + Ex + F)²,
we have the following:
(A²x⁴ + ABx³ + ACx² + ABx³ + B²x² + BCx + ACx² + BCx + C²) -
(D²x⁴ + DEx³ + DFx² + DEx³ + E²x² + EFx + DFx² + EFx + F²) = 0, so
(A²x⁴ + 2ABx³ + 2ACx² + B²x² + 2BCx + C²) - (D²x⁴ + 2DEx³ + 2DFx² + E²x² + 2EFx + F²) = 0,
so A²x⁴ - D²x⁴ + 2ABx³ - 2DEx³ + 2ACx² + B²x² + ACx² - 2DFx² - E²x²
+ 2BCx - 2EFx + C² - F² = 0, so
(A² - D²) * x⁴ + (2AB - 2DE) * x³ + (2AC + B² - 2DF - E²) * x²
+ (2BC - 2EF) * x + (C² - F²) = 0, so
we now have the following 5 equations between our given constants and our new constants:
a = A² - D², b = 2AB - 2DE, c = 2AC - 2DF + B² - E², d = 2BC - 2EF, and e = C² - F².
So, now we can solve them for our new constants, namely A, B, C, D, E, and F:
a + D² = A², b + 2DE = 2AB, c + 2DF = 2AC + B² - E², d + 2EF = 2BC, and e + F² = C²
√(a+D²) = A, (b+2DE)/(2A) = B, 2DF = 2AC + B² - E² - c, 2EF = 2BC - d, and √(e+F²) = C
A=√(a+D²), B=(b+2DE)/(2A), D=(2AC+B²-E²-c)/(2F), E=(2BC-d)/(2F), and C=√(e+F²)
A=√(a+D²), B=(b+2DE)/(2A), C=√(e+F²), D=(2AC+B²-E²-c)/(2F), and E=(2BC-d)/(2F)
Right now, we have B and E in terms of each other, so let's plug in E into B:
B = ( b + 2D [(2BC-d) / (2F)] ) / (2A), and then solve for B:
B = ( b + (D/F) * (2BC-d) ) / (2A) = ( b + 2BCD/F - dD/F ) / (2A)
2AB = b + 2BCD/F - dD/F, so 2AB - 2BCD/F = b - dD/F, so B * (2A - 2CD/F) = b - dD/F,
so B = (b - dD/F) / (2A - 2CD/F) = (bF - dD) / (2AF - 2CD), so now:
A=√(a+D²), B=(bF-dD)/(2AF-2CD), C=√(e+F²), D=(2AC+B²-E²-c)/(2F), and E=(2BC-d)/(2F),
so now if we let F = 1, then we have the following:
A=√(a+D²), B=(b*1-dD)/(2A*1-2CD), C=√(e+1²), D=(2AC+B²-E²-c)/(2*1), and E=(2BC-d)/(2*1),
so A=√(a+D²), B=(b-dD)/(2A-2CD), C=√(e+1), D=(2AC+B²-E²-c)/2, and E=(2BC-d)/2.
Now that we have our new constants in terms of our given constants, we have the following:
(Ax² + Bx + C)² - (Dx² + Ex + F)² = 0
(Ax² + Bx + C)² - (Dx² + Ex + 1)² = 0
(Ax² + Bx + C)² = (Dx² + Ex + 1)²
Ax² + Bx + C = +-(Dx² + Ex + 1)
Ax² + Bx + C = +(Dx² + Ex + 1) or Ax² + Bx + C = -(Dx² + Ex + 1)
Ax² + Bx + C - Dx² - Ex - 1 = 0 or Ax² + Bx + C + Dx² + Ex + 1 = 0
(A-D) * x² + (B-E) * x + (C-1) = 0 or (A+D) * x² + (B+E) * x + (C+1) = 0.
Using the quadratic formula, we now have the following:
x = [ -(B-E) +- √( (B-E)² - 4*(A-D)*(C-1) ) ] / [ 2*(A-D) ] or
x = [ -(B+E) +- √( (B+E)² - 4*(A+D)*(C+1) ) ] / [ 2*(A+D) ].
# *** Cubic formula proof ***:
Given ax³ + bx² + cx + d = 0 = `b`x³ + `c`x² + `d`x + `e` =
0 + `b`x³ + `c`x² + `d`x + `e` = 0x⁴ + `b`x³ + `c`x² + `d`x + `e` =
`a`x⁴ + `b`x³ + `c`x² + `d`x + `e` = (Ax² + Bx + C)² - (Dx² + Ex + F)²,
a = 0, and b =/= 0, we have the following, as seen above, and more below:
`a` = A²-D², `b` = 2AB-2DE, `c` = 2AC - 2DF + B²- E², `d` = 2BC-2EF, and `e` = C²-F²,
so, with F=1, A=√(`a`+D²), B=(`b`-`d`D)/(2A-2CD), C=√(`e`+1), D=(2AC+B²-E²-`c`)/2,
and E=(2BC-`d`)/2, so, with `a`=0, so A=√(0+D²), `b`=a, `c`=b, `d`=c, and `e`=d, we have:
A=√(D²), B=(a-cD)/(2A-2CD), C=√(d+1), D=(2AC+B²-E²-b)/2, and E=(2BC-c)/2
Using the quadratic formula yet again, like last time, we now have the following:
x = [ -(B-E) +- √( (B-E)² - 4*(A-D)*(C-1) ) ] / [ 2*(A-D) ] or
x = [ -(B+E) +- √( (B+E)² - 4*(A+D)*(C+1) ) ] / [ 2*(A+D) ],
but this time we must be careful, since A-D or A+D is = to 0.
