Question 986986
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A quadratic equation is any equation of the form *[tex \Large ax^2\ +\ bx\ +\ c\ =\ 0] where  *[tex \Large a\ \not{=}\ 0].  For your purposes at this point in your study of mathematics, assume that all of the coefficients are real numbers.  However, *[tex \Large y\ =\ ax^2\ +\ bx\ +\ c] is a quadratic <i>equation</i> if and only if *[tex \Large y] is equal to some specific value such that with appropriate algebraic manipulation the equation can be  arranged in to the *[tex \Large ax^2\ +\ bx\ +\ c\ =\ 0] form.  Without such a restriction on the value of *[tex \Large y], *[tex \Large y\ =\ ax^2\ +\ bx\ +\ c] is a quadratic <i>function</i>.


Quadratic <i>functions</i> cannot be "solved" in the commonly accepted sense of finding specific numbers that make the equation a true statement.  The only thing you can say about the solution set of *[tex \Large y\ =\ ax^2\ +\ bx\ +\ c] is that it is *[tex \Large \left\{(x,y)\ \in\ \mathbb{R}^2\,:\,y\ =\ ax^2\ +\ bx\ +\ c\right\}].  Trivial in the extreme.  It is only when you specify a particular value for *[tex \Large y] that the function becomes a quadratic equation that is solvable for a particular pair of values of *[tex \Large x]


In fact, <i>all</i> quadratic equations can be reduced to the product of a pair of factors, although they may not necessarily be rational or even real.  However, unless the factors are rational, finding the factors by ordinary factorization means may be insurmountably challenging.  All of this to say that you stated this question incorrectly. What you are really being asked is if the quadratic equation in question is factorable <i>over the rationals</i>.  Fortunately, there is a simple way to tell:  The quadratic is factorable over the rationals if and only if *[tex \Large b^2\ -\ 4ac] is a perfect square.  In the case of your example, *[tex \Large 49\ -\ 48\ =\ 1] and *[tex \Large 1] is a perfect square.  Therefore the quadratic trinomial *[tex \Large x^2\ +\ 7x\ +\ 12] is factorable over the rationals.  Therefore, *[tex \Large x^2\ +\ 7x\ +\ 12\ =\ 0] can be solved by factoring. Hint: 3 + 4 = 7 and 3 * 4 = 12.


The quadratic formula is:


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


Plug in your numbers and do the arithmetic.


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

*[tex \Large \ \
*[tex \LARGE \ \ \ \ \ \ \ \ \ \