Quadratic equation
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The quadratic equation is a formula for finding the roots of a second-degree polynomial. Any second-degree polynomial in the variable Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle x} will be of the form
where , , and are real constants and is not zero (if it was, the polynomial would only be first-degree). In general the constant coefficients , , and and the variable can be complex, however the quadratic equation is most frequently applied and learnt in the case of real coefficients and a real variable. This corresponds to a parabola, and the roots are then the values of at which the parabola crosses the Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle x} -axis.
This means that the roots of the polynomial are the particular values of Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle x} for which the polynomial equals zero. 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 might not even be real even when the coefficients and variable are. If we call the roots Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle x_+} and Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle x_-} then what we are saying is that
- Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle ax_\pm^2+bx_\pm+c=0\ .}
This is where the quadratic equation comes in. It tells us that the solutions Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle x_+} and Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle x_-} can always be found from the equation
- Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle x_\pm=\frac{-b\pm\sqrt{b^2-4ac}}{2a}\ .}
Looking at the above result it is clear that the quantity Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \Delta\equiv b^2-4ac} (called the discriminant) is of interest, for two reasons. First it is part of the quantity that is either plus or minus depending on whether you are looking at the root Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle x_+} or Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle x_-} . 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.
Three possibilities
Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \Delta>0}
Here we have two distinct real roots, since the square root of Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \Delta} will also be real and greater than zero, meaning that we have
- Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle x_+=\frac{-b+\sqrt{b^2-4ac}}{2a}}
and
- Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle 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
- Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle ax^2+bx+c = \left(x-x_+\right)\left(x-x_-\right)\ .}
Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \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
- Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle x_0=\frac{-b}{2a}}
and the polynomial can be expressed as
- Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle ax^2+bx+c = \left(x-x_0\right)^2\ .}
This case occurs when the parabola described by the polynomial just touches the Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle x} -axis at exactly one point, the red dot shown in Figure 2.
Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \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 Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle x} -axis (or the highest point is below it, if Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle 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 Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle x_\pm} are in fact roots to the polynomial above is to substitute them into the equation
- Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \begin{align}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\pm2b\sqrt{b^2-4ac}+b^2-4ac\right)-\frac{b^2\pm b\sqrt{b^2-4ac}}{2a}+c \\ &=\frac{b^2\pm b\sqrt{b^2-4ac}}{2a}-c-\frac{b^2\pm b\sqrt{b^2-4ac}}{2a}+c \\ &=0\ , \end{align} }
as desired. Notice that this proof is valid even in the case where Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \Delta} is less than zero.