Question 201614
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<b><i>How do you know if a quadratic equation will have one, two, or no solutions?</i></b>.


The short answer is, it is easy:  They all have two solutions.  The Fundamental Theorem of Algebra guarantees it.  Now if what you really meant to ask is: How do you know if a quadratic equation will have one, two, or no solutions <b><i>over the real numbers</i></b>, then read on.


A quadratic equation of the form:


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


has a general solution:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ x = \frac{-b \pm sqrt{b^2 - 4ac}}{2a} ]


The expression under the radical in the general solution, namely *[tex \Large b^2 - 4ac] is called the discriminant.  The discriminant can be evaluated to determine the character of the solutions of a quadratic equation, thus:


If *[tex \Large \Delta = b^2 - 4ac], then


if *[tex \Large \Delta > 0], then the quadratic has two distinct real number roots.  Furthermore, if *[tex \Large \Delta] is a perfect square number, then the roots will be rational, otherwise the roots of the equation will be a conjugate pair of irrational numbers of the form *[tex \Large \alpha \pm \beta\sqrt{\gamma}] where *[tex \Large \alpha,\,\beta,\,\gamma\ \in\ Q,\ \beta\ \neq\ 0,\ \gamma\ \neq\ 0]


if *[tex \Large \Delta = 0], then the quadratic has a single real number root with a multiplicity of 2.  In this case the quadratic is a perfect square having two factors: *[tex \Large (x - \alpha)(x - \alpha)], hence *[tex \Large x = \alpha] is the root, and the multiplicity of 2 comes from the fact that there are two identical factors.


if *[tex \Large \Delta < 0], then the quadratic has no real number solutions.  It has a conjugate pair of complex roots of the form *[tex \Large \alpha \pm \beta i] where *[tex \Large \alpha,\,\beta\ \in\ R,\ \beta\ \neq\ 0] and *[tex \Large i] is the imaginary number defined by *[tex \Large i^2 = -1]


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<b><i>How do you find a quadratic equation if you are only given the solution?</i></b>


Note, that should read "solution<b><i>s</i></b>"


If the solutions of a quadratic equation are *[tex \Large x = \alpha] and *[tex \Large x = \beta], then you can say:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ x - \alpha = 0] and


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ x - \beta = 0], therefore


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ (x - \alpha)(x - \beta) = 0] and finally,


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ x^2 -(\alpha+\beta)x + \alpha\beta = 0]


Hence deriving the quadratic equation from the solutions.  Note that if you are only given one number as a solution, then you have three possibilities.  The first is that you may have a rational number, in which case you have two identical factors that become a perfect square trinomial: *[tex \LARGE \ \ \ \ \ \ \ \ \ \ (x - \alpha)(x - \alpha) = 0 \ \ \Rightarrow\ \ x^2 - 2\alpha x + \alpha^2 = 0].  The second possibility is that you are given an irrational number, such as *[tex \Large \alpha + \beta\sqrt{\gamma}] with the possibility that *[tex \Large \alpha = 0].  In this case, the second root is the conjugate of the given root, namely *[tex \Large \alpha - \beta\sqrt{\gamma}].  The third possibility is that you are given a complex number root, *[tex \Large \alpha + \beta i].  Again, the second root is the conjugate: *[tex \Large \alpha - \beta i]


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<b><i>Is it possible to have different quadratic equations with the same solution?</i></b>


Well, that depends on what you mean by 'different'.  For a given *[tex \Large a], *[tex \Large b], and *[tex \Large c], *[tex \Large ax^2 + bx + c = 0] is equivalent to, and has the same roots as *[tex \Large kax^2 + kbx + kc = 0] for all *[tex \Large k\ \in Q] (actually it is true for all *[tex \Large k\ \in C] but let's stick with rational coefficients for your purposes).  Since the <b><i>equations</i></b> are equivalent, it is a stretch, for me anyway, to call them 'different.'  However, what is true is that the analogous <b><i>functions</i></b>, i.e. *[tex \Large f_i(x) = k_iax^2 + k_ibx + k_ic] are very different in that they all have different graphs.


So while it is true that *[tex \Large f(x) = x^2 - 6x + 5] is different than *[tex \Large g(x) = 2x^2 - 12x + 10] (see graphs below), the equations *[tex \Large x^2 - 6x + 5 = 0] and *[tex \Large 2x^2 - 12x + 10 = 0] are equivalent because you can multiply the second one by *[tex \Large \frac{1}{2}] to obtain the first.  Note that the graphs intersect the *[tex \Large x]-axis at the same points:


{{{drawing(
500, 500, -10, 10, -10, 10,
grid(1),
graph(
500, 500, -10, 10, -10, 10,
x^2 - 6x + 5,
2x^2 - 12x + 10
))}}}


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There should be enough information above for you to come up with a set of solutions for a quadratic for your classmates, particularly in the answer to the second question.


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