quadratic formulaQuadraticFormula2001-10-15 22:08:562007-11-02 03:46:19TheoremAlgebraPolynomial\usepackage{graphicx}
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\newcommand{\figuraex}[2]{\begin{center}\includegraphics[#2]{#1}\end{center}}The roots of the quadratic equation
\[
ax^2+bx+c=0\qquad{a,b,c\in\R,a\neq 0}
\]
are given by the formula
\[
x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}.
\]
The number $\Delta=b^2-4ac$ is called the \emph{discriminant} of the equation.
If $\Delta>0$, there are two different real roots,
if $\Delta=0$ there is a single real root,
and if $\Delta<0$ there are no real roots (but two different complex roots).
Let's work a few examples.
First, consider $2x^2-14x+24=0$.
Here $a=2$, $b=-14$, and $c=24$.
Substituting in the formula gives us
\[
x=\frac{14\pm \sqrt{(-14)^2-4\cdot2\cdot24}}{2\cdot 2}
=\frac{14\pm\sqrt{4}}{4}
=\frac{14\pm2}{4}
=\frac{7\pm1}{2}.
\]
So we have two solutions (depending on whether we take the sign $+$ or $-$):
$x=\frac{8}{2}=4$ and $x=\frac{6}{2}=3$.
Now we will solve $x^2-x-1=0$.
Here $a=1$, $b=-1$, and $c=-1$, so
\[
x=\frac{1\pm\sqrt{(-1)^2-4(1)(-1)}}{2}
=\frac{1\pm{\sqrt{5}}}{2},
\]
and the solutions are $x=\frac{1+\sqrt{5}}{2}$ and $x=\frac{1-\sqrt{5}}{2}$.