Solving a Quartic Equation with Parameters

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[allanmarder]

Syber,
While I liked both methods for solving this problem ( I couldn't think of even 1) I prefer problems where the solution method is more general and applicable to a variety of problems. This problem depends heavily on a "nice factorable equation" or a perfect square lurking in the background. In general, solving a quartic polynomial
with a parameter is not going to give a "nice" answer or even one where the x value can be isolated. Just my thoughts but as I said I liked the solutions!

[yoav]

Nice💯! 2 problems in one, solve the equation and also factor the expression😃

[mcwulf]

Like the second method but needs some trial and error to spot it (which I didn't)

[MrGeorge]

In fact this is easily solvable just starting with either (x² + bx +a) (x² - bx - a + 1) or (x² + bx - a) (x² - bx + a - 1). It can't be ±1 and ±(a - a²) for the constant terms as there are no a² in the coefficients for the x terms.

Eventually (x² + bx - a) (x² - bx + a - 1) with b = -1 solves the problem.

[cicik]

just solve it for a, descriminant will be perfect square and so on