equations A quartic equation where a3 and a1 are equal to 0 takes the form and thus is a biquadratic equation, which is easy to solve: let , so our equation turns to which is a simple quadratic equation, whose solutions are easily found using the quadratic formula: When we've solved it (i.e. found these two z values), we can extract x from them If either of the z solutions were negative or complex numbers, then some of the x solutions are complex numbers. Quasi-symmetric equations Steps: Divide by x 2. Use variable change z = x + m/x. Multiple roots If the quartic has a double root, it can be found by taking the polynomial greatest common divisor with its derivative. Then they can be divided out and the resulting quadratic equation solved. The general case To begin, the quartic must first be converted to a depressed quartic. Converting to a depressed quartic Let be the general quartic equation which it is desired to solve. Divide both sides by , The first step, if is not already zero, should be to eliminate the 3 term. To do this, change variables from to , such that Then Expanding the powers of the binomials produces Collecting the same powers of u yields Now rename the coefficients of . Let The resulting equation is which is a depressed quartic equation. If then we have the special case of a biquadratic equation, which is easily solved, as explained above. Note that the general solution, given below, will not work for the special case The equation must be solved as a biquadratic. In either case, once the depressed quartic is solved for , substituting those values into produces the values for that solve the original quartic. Solving a depressed quartic when ≠ 0 After converting to a depressed quartic equation and excluding the special case = 0, which is solved as a biquadratic, we assume from here on that We will separate the terms left and right as and add in terms to both sides which make them both into perfect squares. Let be any solution of this cubic equation: Then (since ≠ 0) so we may divide by it, giving Then Subtracting, we get the difference of two squares which is the product of the sum and difference of their roots which can be solved by applying the quadratic formula to each of