The
quadratic formula expresses the explicit solution(s)
x to the
quadratic equation
- <math>ax^2+bx+c=0</math>
in terms of the coefficients
a,
b and
c, which are assumed to be
real (but see below for generalizations) with
a being non-zero. These solutions are also called the
roots of the equation. The formula reads
- <math>
x=\frac{-b \pm \sqrt {b^2-4ac}}{2a}
</math>
The term b^{2} – 4ac is called the discriminant of the quadratic equation, because it discriminates between three qualitatively different cases:
- If the discriminant is zero, then there is a repeated solution x, and this solution is real. (Geometrically, this means that the parabola described by the quadratic equation touches the x-axis in a single point.)
- If the discriminant is positive, then there are two different solutions x, both of which are real. (Geometrically, this means that the parabola intersects the x-axis in two points.)
- If the discriminant is negative, then there are two different solutions x, both of which are complex numbers. The two solutions are complex conjugates of each other. (In this case, the parabola does not intersect the x-axis at all.)
Note that when computing roots numerically, the usual form of the quadratic formula is not ideal. See Loss of significance for details.
Derivation
The quadratic formula is derived by the method of completing the square[?].
- <math>ax^2+bx+c=0</math>
Dividing our quadratic equation by a, we have
- <math>
x^2 + \left( \frac{b}{a} \right) x + \frac{c}{a}=0
</math>
which is equivalent to
- <math>x^2+\frac{b}{a}x=-\frac{c}{a}.</math>
The equation is now in a form in which we can conveniently complete the square[?]. To "complete the square" is to add a constant (i.e., in this case, a quantity that does not depend on x) to the expression to the left of "=", that will make it a perfect square trinomial of the form x^{2} + 2xy + y^{2}. Since "2xy" in this case is (b/a)x, we must have y = b/(2a), so we add the square of b/(2a) to both sides, getting
- <math>x^2+\frac{b}{a}x+\frac{b^2}{4a^2}=-\frac{c}{a}+\frac{b^2}{4a^2}.</math>
The left side is now a perfect square; it is the square of (x + b/(2a)). The right side can be written as a single fraction; the common denominator is 4a^{2}. We get
- <math>\left(x+\frac{b}{2a}\right)^2=\frac{-4ac+b^2}{4a^2}=\frac{b^2-4ac}{4a^2}.</math>
Taking square roots of both sides yields
- <math>x+\frac{b}{2a}=\frac{\pm\sqrt{b^2-4ac}}{2a}.</math>
Subtracting b/(2a) from both sides, we get
- <math>x=\frac{-b}{2a}\pm\frac{\sqrt{b^2-4ac}}{2a}=\frac{-b\pm\sqrt{b^2-4ac}}{2a}.</math>
Generalizations
The formula and its proof remain correct if the coefficients a, b and c are complex numbers, or more generally members of any field whose characteristic is not 2. (In a field of characteristic 2, the element 2a is zero and it is impossible to divide by it.)
The symbol
- <math>\pm \sqrt {b^2-4ac}</math>
in the formula should be understood as "either of those element(s) of the field whose square equals
b^{2} - 4
ac". In some fields, some elements have no square roots and some have two; only zero has just one square root, except in fields of characteristic 2.
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