Let's consider the equation 3x3 + 4x2 + 9x + 2 = 0 (Or any other cubic equation). If we divide everything by 3x3, we get 1 + 4/3x + 1/3x2 + 2/3x3 = 0. Now we can subtract 1 from each side and get 4/3x + 1/3x2 + 2/3x3 = -1. If we raise everything to a power of -1, we get 3x/4 + 3x2 + 3x3/2 = -1. Now we add one and get 3x/4 + 3x2 + 3x3/2 + 1 = 0. Rearranging terms and multiplying by 2, We get 3x3 + 6x2 + 3x/2 + 2 = 0. We now have two expressions equal to zero, so we can set them equal to each other. 3x3 + 6x2 + 3x/2 + 2 = 3x3 + 4x2 + 9x + 2. The 3x3 and 2 terms cancel out, leaving 6x2 + 3x/2 = 4x2 +9x. Moving all the terms to right the right side makes the new equation -2x2 + 6x = 0. The solutions to this problem are 0 and 3, but 0 and 3 aren't solutions to the original problem. What am I doing wrong and how do I solve a cubic equation? I know there is a formula but is there a faster way?
It's at the point where you "raise everything to the power minus 1 " that you are wrong, that simply doesn't work.
Consider for instance
\(\displaystyle \frac{1}{6} + \frac{1}{3}=\frac{1}{2}\),
raise everything to the power -1 and you get 6 + 3 = 2, which is clearly nonsense.
Cubic equations with real coefficients have three roots, all three roots could be real, or there could be one real root and two complex roots, (and if that's the case the complex roots will form a conjugate pair). (There can't be one or three complex roots, it's two or none).
To solve the equation, you might get lucky and spot a/the real root, or failing that you could use a numerical method to find its (approximate) value. Dividing that factor out then reduces the problem to solving a quadratic.
Algebraic methods for solving the equation will involve the use of complex numbers, look at Cardin's method for example.
Tiggsy
It has no "real" solution but "complex" solution!!. If you wish to see how to solve it for its complex roots, then let us know.
Yes, there is at least ONE "real" solution, using Newton-Raphson method:
x≈ -0.243810611734410
It's at the point where you "raise everything to the power minus 1 " that you are wrong, that simply doesn't work.
Consider for instance
\(\displaystyle \frac{1}{6} + \frac{1}{3}=\frac{1}{2}\),
raise everything to the power -1 and you get 6 + 3 = 2, which is clearly nonsense.
Cubic equations with real coefficients have three roots, all three roots could be real, or there could be one real root and two complex roots, (and if that's the case the complex roots will form a conjugate pair). (There can't be one or three complex roots, it's two or none).
To solve the equation, you might get lucky and spot a/the real root, or failing that you could use a numerical method to find its (approximate) value. Dividing that factor out then reduces the problem to solving a quadratic.
Algebraic methods for solving the equation will involve the use of complex numbers, look at Cardin's method for example.
Tiggsy
