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quadratic formula (Theorem)

The roots of the quadratic equation

\begin{displaymath} ax^2+bx+c=0\qquad{a,b,c\in\mathbbmss{R},a\neq 0} \end{displaymath}

are given by the formula
\begin{displaymath} x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}. \end{displaymath}

The number $\Delta=b^2-4ac$ is called the 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

\begin{displaymath} 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}. \end{displaymath}

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

\begin{displaymath} x=\frac{1\pm\sqrt{(-1)^2-4(1)(-1)}}{2} =\frac{1\pm{\sqrt{5}}}{2}, \end{displaymath}

and the solutions are $x=\frac{1+\sqrt{5}}{2}$ and $x=\frac{1-\sqrt{5}}{2}$.



"quadratic formula" is owned by yark. [ full author list (2) | owner history (2) ]
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See Also: derivation of quadratic formula, quadratic inequality, quadratic equation in $\mathbb{C}$, conjugated roots of equation, quadratic congruence

Keywords:  Algebra, Polynomial

Attachments:
derivation of quadratic formula (Proof) by mathcam
properties of quadratic equation (Result) by pahio
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Cross-references: solutions, complex roots, real, equation, discriminant, quadratic equation, roots
There are 16 references to this entry.

This is version 8 of quadratic formula, born on 2001-10-15, modified 2007-11-02.
Object id is 227, canonical name is QuadraticFormula.
Accessed 32951 times total.

Classification:
AMS MSC: 12D10 (Field theory and polynomials :: Real and complex fields :: Polynomials: location of zeros )
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