Question 340006
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The first part of your question is easy.  There are ALWAYS two solutions, if you count the multiplicity.  The Fundamental Theorem of Algebra says so.  However, evaluate the discriminant to determine the nature of the roots.


For any quadratic polynomial equation of the form:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ ax^2\ +\ bx\ +\ c\ = 0]


Find the Discriminant, *[tex \LARGE \Delta\ =\ b^2\ -\ 4ac] and evaluate the nature of the roots as follows:


No calculation quick look:  If the signs on *[tex \Large a] and *[tex \Large c] are opposite, then *[tex \LARGE \Delta > 0] guaranteed.


*[tex \LARGE \Delta > 0 \ \ \Rightarrow\ \] Two real and unequal roots. If *[tex \LARGE \Delta] is a perfect square, the quadratic factors over *[tex \LARGE \mathbb{Q}].


*[tex \LARGE \Delta = 0 \ \ \Rightarrow\ \] One real root with a multiplicity of two.  That is to say that the trinomial is a perfect square and has two identical factors.


*[tex \LARGE \Delta < 0 \ \ \Rightarrow\ \] A conjugate pair of complex roots of the form *[tex \LARGE a \pm bi] where *[tex \LARGE i] is the imaginary number defined by *[tex \LARGE i^2 = -1]


+----------+----------+----------+----------+----------+----------+----------+


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ x\ =\ \alpha]


is a solution of the quadratic equation:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ ax^2\ +\ bx\ +\ c\ = 0]


if and only if


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ x\ -\ \alpha]


is a factor of the quadratic polynomial:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ ax^2\ +\ bx\ +\ c]


So, just take your given solution, whatever it is, and form a binomial factor by subtracting from x.  Do this for each of your solutions.  If you are told there is only one solution, say, out loud, "Ok, there is a pair of identical solutions" and create two identical factors.  If you are given one irrational solution, remember that irrational solutions always come in conjugate pairs.  If you are given *[tex \Large \varphi\ +\ \sigma\sqrt{\rho}] as a solution, then you know that *[tex \Large \varphi\ -\ \sigma\sqrt{\rho}] must also be a solution. The conjugate trick works for complex roots (the ones that involve *[tex \Large i] that you get when *[tex \LARGE \Delta < 0]) also.


Once you have your two factors, multiply them together using our old friend FOIL, and set the result equal to zero.  That will give you one possible quadratic equation with the given roots.


+----------+----------+----------+----------+----------+----------+----------+


It is most assuredly possible.  In fact, for any pair of solutions, real or otherwise, there is a set of quadratic equations with infinite elements.


If *[tex \LARGE \{\alpha,\beta\}] is the solution set of the quadratic equation


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ ax^2\ +\ bx\ +\ c\ = 0]


Then *[tex \LARGE \{\alpha,\beta\}] is the solution set of every quadratic equation of the form:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ akx^2\ +\ bkx\ +\ ck\ = 0]


where *[tex \LARGE k\ \in\ \mathbb{R}]


John
*[tex \LARGE e^{i\pi} + 1 = 0]
My calculator said it, I believe it, that settles it
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