# On the number of possible solutions for a quadratic equation.

Solving a quadratic equation will yield two roots:

$$\frac{-\sqrt{b^2-4 a c}-b} {2 a}$$

and:

$$\frac{\sqrt{b^2-4a c}-b}{2 a}$$

And I've been taught to answer it like:

$$\frac{\pm\sqrt{b^2-4a c}- b}{2 a}$$

Why does it yields only two solutions? Aren't there infinite solutions for that? Is there a proof on the number of possible solutions for a quadratic equation?

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Uhhhh...no. For constant (real number) values of $a$, $b$, and $c$, the polynomial equation $ax^2 + bx + c = 0$ has at most two real roots--and has exactly two complex roots. – user5137 Aug 25 '12 at 2:07
Your general solution is wrong. It must be $$\frac{-b\pm\sqrt {b^2-4ac}}{2a}=\frac{\pm\sqrt{b^2-4ac}-b}{2a}$$ – DonAntonio Aug 25 '12 at 2:11
As Gauss said... – J. M. Aug 25 '12 at 2:15
There are two solutions because $a,b,c$ are all given to you. You aren't choosing them: by the time you're applying this formula you should know what they are. So the only thing you have to choose is the $\pm$, giving you two solutions. – Robert Mastragostino Aug 25 '12 at 3:38
Ah, I see. In which case, as others have shown, it's because there are two values of $x$ for a given $x^2$ (say, $2$ and $-2$ for $4$), and so when you're solving the equation you have to consider both. You get two for the price of one. – Robert Mastragostino Aug 28 '12 at 15:36

One can see that there are only two solutions from the way it is solved. If $ax^2+bx+c=0$ with $a\neq 0$, it follows that $ax^2+bx=-c$, so that $x^2+\frac{b}{a}x=-\frac{c}{a}$. Trying to complete the square, one obtains $(x+\frac{b}{2a})^2=x^2+\frac{b}{a}x+(\frac{b}{2a})^2=(\frac{b}{2a})^2-\frac{c}{a}=\frac{b^2-4ac}{4a^2}$. Taking square roots, the result follows.

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Think about the graph of $y = ax^2 + bx + c$ (where $a \ne 0$). It's a parabola. It may be concave or convex, depending on whether $a>0$ or $a<0$, but it's still a parabola. The roots of $ax^2 + bx + c = 0$ are the places where $y=0$; in other words, the places where this parabola crosses the $x$-axis. Usually, there will be either two crossings, or none. There's a special case when the parabola just touches the axis. In this case, the quadratic has two equal roots.

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Any polynomial equation of the form $$a_nx^n+a_{n-1}x^{n-1}+...+a_1x+a_0=0\,\,,\,\,a_i\in\Bbb C$$ has exactly n solutions, counting multiplicities (this much is true over more general fields, of course)

Thus, a quadratic over any field has two roots over any algebraically closed field. If you're working over the reals though, then sometimes there are no real roots (solutions) to a quadratic, but there always be two solutions on the complex field.

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This is to do with the fact that every polynomial over the complex numbers has a "unique factorization". The proof is based on Euclidean algorithm for polynomials.

It is not the case anymore when you allow solutions outside the domain of complex numbers. For example, there are infinitely many quaternions that satisfy $x^2 = -1$.

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A more important question: Why should there be infinitely many solutions?

The roots/solutions of a quadratic equation $ax^2 + bx + c = 0$ are given by looking at when the graphs $y = ax^2 + bx + c$ cuts the $x$-axis. Remember that $a$, $b$ and $c$ are all fixed numbers, e.g. $y = x^2 + 2x + 1.$ Ask yourself this: Why should there be infinity many values of $x$ for which $x^2 + 2x + 1 = 0$? In fact, there is only one single value of $x$ for which $x^2 + 2x + 1 = 0$. This is because $x^2 + 2x + 1 = (x+1)^2$ and so $x = -1$ is the only solution. Find me another value of $x$ for which $x^2 + 2x + 1 = 0$ and I will give you US$1,000,000. Draw yourself some sketches. Draw the one-parameter family of parabolae given by$y = x^2 - k$where$k$is a number we're going to play with. When$k > 0$, the parabola cuts the$x$-axis at two distinct points:$x = \pm \sqrt{k}$. When$k = 0$, the parabola cuts the$x$-axis at one point:$x = 0$. When$k < 0$, the parabola misses the$x$-axis altogether. - "Find another value to$x\$..." Sorry, TD4I. – Voyska Aug 25 '12 at 20:00