In mathematics, a quadratic equation is a polynomial equation of the second degree. The generalized form is

[itex]ax^2+bx+c=y\mbox{ where }a\ne 0.[itex]

The letters a, b and c are called coefficients: a is the coefficient of x2, b is the coefficient of x, and c is the free term or constant.

A quadratic equation with real or complex coefficients has two complex roots (i.e., solutions for x when y = 0) usually denoted as [itex]x_1[itex] and [itex]x_2[itex], although the two roots may be equal. These roots can be computed using the quadratic formula.

Higher-degree equations may be quadratic in form, such as:

[itex]2x^6+3x^3+5=0[itex].

Note that the highest exponent is twice the value of the exponent of the middle term. This equation may be resolved directly or with a simple substitution, using the methods that are available for the quadratic, such as factoring (also called factorising), the quadratic formula, or completing the square.

 Contents

The quadratic formula explicitly gives the solutions of a quadratic equation in terms of the coefficients a, b and c, which we temporarily assume 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

[itex]

x_{1,2}=\frac{-b \pm \sqrt {b^2-4ac\ }}{2a}. [itex]

An alternative form sometimes encountered is given by

[itex]

x_{1,2}=\frac{2c}{-b \pm \sqrt {b^2-4ac\ }}. [itex] Template:ExampleSidebar The term b2 − 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. Furthermore, if the discriminant is a perfect square, the roots are rational numbers -- in other cases they may be quadratic irrationals.
• 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.

[itex]ax^2+bx+c=0[itex]

Dividing our quadratic equation by a, we have

[itex]

x^2 + \frac{b}{a} x + \frac{c}{a}=0 [itex]

which is equivalent to

[itex]x^2+\frac{b}{a}x=-\frac{c}{a}.[itex]

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 x2 + 2xy + y2. 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

[itex]x^2+\frac{b}{a}x+\frac{b^2}{4a^2}=-\frac{c}{a}+\frac{b^2}{4a^2}.[itex]

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 4a2. We get

[itex]\left(x+\frac{b}{2a}\right)^2=\frac{b^2-4ac}{4a^2}.[itex]

Taking square roots of both sides yields

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

Subtracting b/(2a) from both sides, we get

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

## 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

[itex]\pm \sqrt {b^2-4ac }[itex]

in the formula should be understood as "either of the two elements whose square is b2 − 4ac, if such elements exist". 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.

## Viète's formulas

Viète's formulas give a simple relation between the roots of a polynomial and its coefficients. In the case of quadratic polynomial, they take the following form:

[itex] x_1 + x_2 = \frac{-b}{a}, [itex]
[itex] x_1 x_2 = \frac{c}{a}. [itex]

## History

The ancient Babylonians (around 400 BC) and Chinese used the method of completing the square to solve quadratic equations with positive roots, but did not have a general formula. Euclid produced a more abstract geometrical method around 300 BC.

The first mathematician known to have used the general algebraic formula, allowing negative as well as positive solutions, was Brahmagupta (India, 7th century). Al-Khwarizmi (Arabia, 11th century) independently developed a set of formulae that worked for positive solutions. Abraham bar Hiyya Ha-Nasi (also known by the Latin name Savasorda) was the first to introduce the complete solution to Europe in his book Liber embadorum.

Sridhara was said to be one of the first mathematicians to give a general rule for solving a quadratic equation. But there has been a dispute over his time. The rule is (as quoted by Bhaskara II):

Multiply both sides of the equation by a known quantity equal to four times the coefficient of the square of the unknown; add to both sides a known quantity equal to the square of the coefficient of the unknown; then take the square root. (http://www-gap.dcs.st-and.ac.uk/~history/Mathematicians/Sridhara.html)

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