Question 251041
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Given no other information than rational number solution, such as your example of -3, you can't create a unique quadratic equation.  However, if you have either a real irrational solution or a complex solution, then you only have to realize that such solutions always come in conjugate pairs.


Let *[tex \Large \alpha] and *[tex \Large \beta] be any rational numbers.  Let *[tex \Large q] be a non-transcendental irrational number.  Then if *[tex \Large \alpha\ \pm\ \beta q] is a root of the quadratic, then *[tex \Large \alpha\ \mp\ \beta q] is also a solution of the quadratic.


Let *[tex \Large \alpha] and *[tex \Large \beta] be any real numbers.  Let *[tex \Large i] be the imaginary number defined by *[tex \Large i^2\ =\ -1].  Then if *[tex \Large \alpha\ \pm\ \beta i] is a root of the quadratic, then *[tex \Large \alpha\ \mp\ \beta i] is also a solution of the quadratic.


On the other hand, if you know another bit of information about the quadratic for which you have a single rational number solution, then you can derive an equation for the quadratic.  For example, we know that the graph of a quadratic of the form *[tex \LARGE f(x)\ =\ y\ =\ ax^2\ +\ bx\ +\ c] is a parabola symmetric to the vertical line that passes through its vertex.  So if the vertex of the parabola is at *[tex \LARGE \left(x_v,\,f(x_v)\right)], and the quadratic has real number roots, then the two roots must be equidistant from the vertex, and therefore equidistant from the intersection of the parabola's axis and the *[tex \LARGE x]-axis, the line *[tex \LARGE x\ =\ x_v].  Using your example of -3 for a root, the distance from the intersection of the parabola axis and the *[tex \LARGE x]-axis is exactly *[tex \LARGE x_v\ -\ (-3)].  So, depending on which side of the vertex -3 exists, you either add or subtract the distance value you just calculated from *[tex \LARGE x_v] to find the other root.


John
*[tex \LARGE e^{i\pi} + 1 = 0]
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