root the term in () is always real, which ensures that the two quadratic equations () have real coefficients. Obtaining alternative solutions the hard way It could happen that one only obtained one solution through the formulae above, because not all four sign patterns are tried for four solutions, and the solution obtained is complex. It may also be the case that one is only looking for a real solution. Let x1 denote the complex solution. If all the original coefficients A, B, C, D and E are real—which should be the case when one desires only real solutions – then there is another complex solution x2 which is the complex conjugate of x1. If the other two roots are denoted as x3 and x4 then the quartic equation can be expressed as but this quartic equation is equivalent to the product of two quadratic equations: and Since then Let so that equation () becomes Also let there be (unknown) variables w and v such that equation () becomes Multiplying equations () and () produces Comparing equation () to the original quartic equation, it can be seen that and Therefore Equation () can be solved for x yielding One of these two solutions should be the desired real solution. Alternative methods Quick and memorable solution from first principles Most textbook solutions of the quartic equation require a magic substitution that is almost impossible to memorize. Here is a way to approach it that makes it easy to understand. The job is done if we can factor the quartic equation into a product of two quadratics. Let By equating coefficients, this results in the following set of simultaneous equations: This is harder to solve than it looks, but if we start again with a depressed quartic where , which can be obtained by substituting for , then , and: It's now easy to eliminate both and by doing the following: If we set , then this equation turns into the cubic equation: which is solved elsewhere. Once you have , then: The symmetries in this solution are easy to see. There are three roots of the cubic, corresponding to the three ways that a quartic can be factored into two quadratics, and choosing positive or negative values of for the square root of merely exchanges the two quadratics with one another. Galois theory and factorization The symmetric group S4 on four