Quadratic equation

In mathematics, or more specifically algebra, a quadratic equation is one involving only polynomials of the second degree. Quadratic equations are a common part of mathematical solutions to real-world problems in a huge variety of situations. Fortunately, then, there exists a simple closed formula for finding the roots of such an equation, the quadratic formula.

Every polynomial equation can be put into the form:

$ax^{2}+bx+c=0\,$

with a, b and c real and $a\not =0$. The quadratic formula specifies the roots of this equation as

$x={\frac {-b\pm {\sqrt {b^{2}-4ac}}}{2a}}\ .$

Here, there are actually two roots being given, although the notation obscures this somewhat. One root is obtained by replacing the $\pm$ appearing in the formula by a + sign, and the other is obtained by replacing it with a - sign.

It is therefore very useful for factoring. The formula is guaranteed to work for all quadratic polynomials, but sometimes the roots will be complex numbers even when every other part of the problem deals only with real numbers.

In this article, we are assuming as above that the coefficients of the quadratic polynomial are real. Quadratic equations can occur involving many other types coefficients, and the quadratic formula, interpreted correctly, can be applied in many of these other situations as well. See the advanced subpage for information about these more general quadratic equations.

The problem

Any real second-degree polynomial in the variable $x$ will be of the form

$ax^{2}+bx+c\,$

where $a$, $b$, and $c$ are real constants and $a$ is not zero (if it was, the polynomial would only be first-degree). A polynomial of this form corresponds to a parabola, and the roots that the quadratic equation will give us are the values of $x$ at which the parabola crosses the $x$-axis. This means that the roots of the polynomial are the particular values of $x$ for which the polynomial equals zero.

The problem that the quadratic formula solves is to find those roots.

The solution

The Fundamental Theorem of Algebra tells us that we should expect there to be two roots for a second-degree polynomial, although they might be equal in some cases, and may not be real even when the coefficients and variable are. If we call the roots $x_{+}$ and $x_{-}$ then what we are saying is that we have the quadratic equation

$ax_{\pm }^{2}+bx_{\pm }+c=0\ .$

This is where the quadratic formula comes in. It tells us that the solutions $x_{+}$ and $x_{-}$ can always be found as

$x_{\pm }={\frac {-b\pm {\sqrt {b^{2}-4ac}}}{2a}}\ .$

Looking at the above result it is clear that the quantity $\Delta =b^{2}-4ac$ (called the discriminant) is of interest, for two reasons. First it is part of the quantity that is either positive or negative depending on whether you are looking at the root $x_{+}$ or $x_{-}$, so it is this quantity that is responsible for whether we have two different roots or not. Second it is under a square root, so we must wonder what happens when it is negative. This gives us three cases to look at.

Figure 1: A parabola with two real roots. This corresponds to a second-degree (real) polynomial with a positive discriminant.

Figure 2: A parabola with two only one real root, but of multiplicity 2. This corresponds to a second-degree (real) polynomial with a discriminant equal to zero.

Figure 3: A parabola with no real roots, as seen by the fact that it does not intersect the horizontal axis. This corresponds to a second-degree (real) polynomial with a negative discriminant.

$\Delta >0$

Here we have two distinct real roots, since the square root of $\Delta$ will also be real and greater than zero, meaning that we have

$x_{+}={\frac {-b+{\sqrt {b^{2}-4ac}}}{2a}}$

and

$x_{-}=x_{\pm }={\frac {-b-{\sqrt {b^{2}-4ac}}}{2a}}\ .$

These can be seen graphically as the two red dots in Figure 1. In this case we can rewrite the polynomial in terms of its roots as

$ax^{2}+bx+c=\left(x-x_{+}\right)\left(x-x_{-}\right)\ ,$

and it is easy to see that indeed if we set $x=x_{+}$ or $x=x_{-}$ then the polynomial well be equal to zero.

$\Delta =0$

Now there is only one distinct root. It is still real, and is said to have a multiplicity of 2. This is because the root is given by

$x_{0}={\frac {-b}{2a}}$

and the polynomial can be expressed as

$ax^{2}+bx+c=\left(x-x_{0}\right)^{2}\ .$

This case occurs when the parabola described by the polynomial just touches the $x$-axis at exactly one point, the red dot shown in Figure 2. Again, setting $x=x_{0}$ in the polynomial makes it vanish.

$\Delta <0$

For negative values of the discriminant there are no real roots to the polynomial. Graphically this corresponds to the situation where the lowest point on the parabola is above the $x$-axis (or the highest point is below it, if $a$ is negative) as shown in Figure 3. In this case the roots still exist, as guaranteed by the fundamental theorem of algebra, but they are complex so cannot be shown on the real number line.

Proof

The simplest way to show that the values $x_{\pm }$ are in fact roots to the polynomial above is to substitute them into the equation

${\begin{aligned}ax_{\pm }^{2}+bx_{\pm }+c&=a\left({\frac {-b\pm {\sqrt {b^{2}-4ac}}}{2a}}\right)^{2}+b{\frac {-b\pm {\sqrt {b^{2}-4ac}}}{2a}}+c\\&={\frac {1}{4a}}\left(b^{2}\mp 2b{\sqrt {b^{2}-4ac}}+b^{2}-4ac\right)-{\frac {b^{2}\mp b{\sqrt {b^{2}-4ac}}}{2a}}+c\\&={\frac {b^{2}\mp b{\sqrt {b^{2}-4ac}}}{2a}}-c-{\frac {b^{2}\mp b{\sqrt {b^{2}-4ac}}}{2a}}+c\\&=0\ ,\end{aligned}}$

as desired. Notice that this proof is valid even in the case where $\Delta$ is less than zero.

Derivation of the quadratic formula

To derive the quadratic formula stated above, we must start with the quadratic equation

$ax^{2}+bx+c=0$

and then complete the square. We start by subtracting $c$ from both sides and then writing the left-hand side of the equation as a complete square plus a constant term,

$\left({\sqrt {a}}x+{\frac {b}{2{\sqrt {a}}}}\right)^{2}-{\frac {b^{2}}{4a}}=-c\ .$

By re-arranging this and taking either the positive or negative value of the square root we can isolate the term containing $x$,

${\sqrt {a}}x+{\frac {b}{2{\sqrt {a}}}}=\pm {\sqrt {\frac {b^{2}-4ac}{4a}}}\quad \implies \quad {\sqrt {a}}x={\frac {-b\pm {\sqrt {b^{2}-4ac}}}{2{\sqrt {a}}}}\ .$

Finally we divide through by the square root of $a$ and have arrived at the quadratic formula,

$x={\frac {-b\pm {\sqrt {b^{2}-4ac}}}{2a}}\ .$