Quadratic equation

Quadratic equation

General form

$x^2+\frac{b}{a}x+\frac{c}{a}=0.$

Derivation of the formula

The quadratic formula can be derived with a simple application of technique of completing the square.Divide the quadratic equation by Template:Math, which is allowed because Template:Math is non-zero:

$x^2+\frac{b}{a}x+\frac{c}{a}=0.$

Subtract Template:Math from both sides of the equation, yielding:

$x^2+\frac{b}{a}x=-\frac{c}{a}.$

The quadratic equation is now in a form to which the method of completing the square can be applied. Thus, add a constant to both sides of the equation such that the left hand side becomes a complete square:

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

which produces:

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

Accordingly, after rearranging the terms on the right hand side to have a common denominator, we obtain:

${(x+\frac{b}{2a})}^2=\frac{b^2-4ac}{4a^2}.$

The square has thus been completed. Taking the square root of both sides yields the following equation:

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

Isolating Template:Math gives the quadratic formula:

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

The plus-minus symbol "±" indicates that both

$x=\frac{-b+\sqrt{b^2-4ac}}{2a}\text{and}x=\frac{-b-\sqrt{b^2-4ac}}{2a}$

are solutions of the quadratic equation. There are many alternatives of this derivation with minor differences, mostly concerning the manipulation of Template:Math.

Some sources, particularly older ones, use alternative parameterizations of the quadratic equation such as Template:Math or Template:Math, where Template:Math has a magnitude one half of the more common one. These result in slightly different forms for the solution, but are otherwise equivalent.

A lesser known quadratic formula, as used in Muller's method, and which can be found from Vieta's formulas, provides the same roots via the equation:

$x=\frac{2c}{-b\mp \sqrt{b^2-4ac}}.$