Quadratic equation

This article is about quadratic equations and solutions. For more general information about quadratic functions, see Quadratic function. For more information about quadratic polynomials, see Quadratic polynomial.

In mathematics, a quadratic equation is a univariate polynomial equation of the second degree. A general quadratic equation can be written in the form

$ax^2+bx+c=0,\,$

where x represents a variable or an unknown, and a, b, and c are constants with a ≠ 0. (If a = 0, the equation is a linear equation.)

The constants a, b, and c are called respectively, the quadratic coefficient, the linear coefficient and the constant term or free term. The term "quadratic" comes from quadratus, which is the Latin word for "square". Quadratic equations can be solved by factoring, completing the square, graphing, Newton's method, and using the quadratic formula (given below).

Plots of real-valued quadratic function ax² + bx + c, varying each coefficient separately

Quadratic formula

A quadratic equation with real or complex coefficients has two solutions, called roots. These two solutions may or may not be distinct, and they may or may not be real.

The roots are given by the quadratic formula

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

where the symbol "±" indicates that both

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

are solutions of the quadratic equation.

Discriminant

Example discriminant signs
■ <0: x²+¹⁄₂
■ =0: −⁴⁄₃x²+⁴⁄₃x−¹⁄₃
■ >0: ³⁄₂x²+¹⁄₂x−⁴⁄₃

In the above formula, the expression underneath the square root sign is called the discriminant of the quadratic equation, and is often represented using an upper case Greek delta, the initial of the Greek word Διακρίνουσα, Diakrínousa, discriminant:

$\Delta = b^2 - 4ac.\,$

A quadratic equation with real coefficients can have either one or two distinct real roots, or two distinct complex roots. In this case the discriminant determines the number and nature of the roots. There are three cases:

• If the discriminant is positive, then there are two distinct roots, both of which are real numbers:

$\frac{-b + \sqrt {\Delta}}{2a} \quad\text{and}\quad \frac{-b - \sqrt {\Delta}}{2a}$

For quadratic equations with integer coefficients, if the discriminant is a perfect square, then the roots are rational numbers—in other cases they may be quadratic irrationals.

• If the discriminant is zero, then there is exactly one distinct real root, sometimes called a double root:

$x = -\frac{b}{2a} . \,\!$

• If the discriminant is negative, then there are no real roots. Rather, there are two distinct (non-real) complex roots, which are complex conjugates of each other:

$\frac{-b}{2a} + i \frac{\sqrt {-\Delta}}{2a}, \quad\text{and}\quad \frac{-b}{2a} - i \frac{\sqrt {-\Delta}}{2a},$

where i is the imaginary unit.

Thus the roots are distinct if and only if the discriminant is non-zero, and the roots are real if and only if the discriminant is non-negative.

Monic form

Dividing the quadratic equation by coefficient a gives the simplified monic form of

$x^2 + px + q =\,0\text{,}$

where p = ^b⁄_a and q = ^c⁄_a. This in turn simplifies the root and discriminant equations somewhat to

$x = \frac{1}{2} \left( -p \pm \sqrt{p^2 - 4q} \right)$

and

$\Delta = p^2 - 4q\,\text{.}$

History

The Babylonian mathematicians, as early as 2000 BC (displayed on Old Babylonian clay tablets) could solve a pair of simultaneous equations of the form:

$x+y=p,\ \ xy=q \$

which are equivalent to the equation:^[1]

$\ x^2+q=px$

The original pair of equations were solved as follows:

1. Form    $\frac{x+y}{2}$
2. Form    $\left(\frac{x+y}{2}\right)^2$
3. Form    $\left(\frac{x+y}{2}\right)^2 - xy$
4. Form    $\sqrt{\left(\frac{x+y}{2}\right)^2 - xy} = \frac{x-y}{2}$      (where x ≥ y is assumed)
5. Find x and y by inspection of the values in (1) and (4).^[2]

There is evidence pushing this back as far as the Ur III dynasty.^[3]

In the Sulba Sutras in ancient India circa 8th century BC quadratic equations of the form ax² = c and ax² + bx = c were explored using geometric methods. Babylonian mathematicians from circa 400 BC and Chinese mathematicians from circa 200 BC used the method of completing the square to solve quadratic equations with positive roots, but did not have a general formula.^{[citation needed]}

Euclid, the Greek mathematician, produced a more abstract geometrical method around 300 BC. Pythagoras and Euclid used a strictly geometric approach, and found a general procedure to solve the quadratic equation.^[4] In his work Arithmetica, the Greek mathematician Diophantus solved the quadratic equation, but giving only one root, even when both roots were positive.^[5]

In 628 AD, Brahmagupta, an Indian mathematician, gave the first explicit (although still not completely general) solution of the quadratic equation

$\ ax^2+bx=c \,$

as follows:

“

To the absolute number multiplied by four times the [coefficient of the] square, add the square of the [coefficient of the] middle term; the square root of the same, less the [coefficient of the] middle term, being divided by twice the [coefficient of the] square is the value. (Brahmasphutasiddhanta (Colebrook translation, 1817, page 346)[2]

”

This is equivalent to:

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

The Bakhshali Manuscript written in India in the 7th century AD contained an algebraic formula for solving quadratic equations, as well as quadratic indeterminate equations (originally of type ax/c = y).

Muhammad ibn Musa al-Khwarizmi (Persia, 9th century), inspired by Brahmagupta, developed a set of formulas that worked for positive solutions.^[4] Al-Khwarizmi goes further in providing a full solution to the general quadratic equation, accepting one or two numerical answers for every quadratic equation, while providing geometric proofs in the process.^[6] He also described the method of completing the square and recognized that the discriminant must be positive,^[7] which was proven by his contemporary 'Abd al-Hamīd ibn Turk (Central Asia, 9th century) who gave geometric figures to prove that if the discriminant is negative, a quadratic equation has no solution.^[8] While al-Khwarizmi himself did not accept negative solutions, later Islamic mathematicians that succeeded him accepted negative solutions,^[9] as well as irrational numbers as solutions.^[10] Abū Kāmil Shujā ibn Aslam (Egypt, 10th century) in particular was the first to accept irrational numbers (often in the form of a square root, cube root or fourth root) as solutions to quadratic equations or as coefficients in an equation.^[11]

The Jewish mathematician Abraham bar Hiyya Ha-Nasi (12th century, Spain) authored the first European book to include the full solution to the general quadratic equation.^[12] His solution was largely based on Al-Khwarizmi's work.^[13] The writing of the Chinese mathematician Yang Hui (1238-1298 AD) represents the first in which quadratic equations with negative coefficients of 'x' appear, although he attributes this to the earlier Liu Yi.

By 1545 Gerolamo Cardano compiled the works related to the quadratic equations. The quadratic formula covering all cases was first obtained by Simon Stevin in 1594.^[14] In 1637 René Descartes published La Géométrie containing the quadratic formula in the form we know today.^[4] The first appearance of the general solution in the modern mathematical literature appeared in a 1896 paper by Henry Heaton.^[15]

Examples of use

Geometry

For the quadratic function:
f (x) = x² − x − 2 = (x + 1)(x − 2) of a real variable x, the x-coordinates of the points where the graph intersects the x-axis, x = −1 and x = 2, are the solutions of the quadratic equation: x² − x − 2 = 0.

The solutions of the quadratic equation

$ax^2+bx+c=0,\,$

are also the roots of the quadratic function:

$f(x) = ax^2+bx+c,\,$

since they are the values of x for which

$f(x) = 0.\,$

If a, b, and c are real numbers and the domain of f is the set of real numbers, then the roots of f are exactly the x-coordinates of the points where the graph touches the x-axis.

It follows from the above that, if the discriminant is positive, the graph touches the x-axis at two points, if zero, the graph touches at one point, and if negative, the graph does not touch the x-axis.

Quadratic factorization

The term

$x - r\,$

is a factor of the polynomial

$ax^2+bx+c, \$

if and only if r is a root of the quadratic equation

$ax^2+bx+c=0. \$

It follows from the quadratic formula that

$ax^2+bx+c = a \left( x - \frac{-b + \sqrt {b^2-4ac}}{2a} \right) \left( x - \frac{-b - \sqrt {b^2-4ac}}{2a} \right).$

In the special case (b² = 4ac) where the quadratic has only one distinct root (i.e. the discriminant is zero), the quadratic polynomial can be factored as

$ax^2+bx+c = a \left( x + \frac{b}{2a} \right)^2.\,\!$

Application to higher-degree equations

Certain higher-degree equations can be brought into quadratic form and solved that way. For example, the 6th-degree equation in x:

$x^6 - 4x^3 + 8 = 0\,$

can be rewritten as:

$(x^3)^2 - 4(x^3) + 8 = 0\,,$

or, equivalently, as a quadratic equation in a new variable u:

$u^2 - 4u + 8 = 0,\,$

where

$u = x^3.\,$

Solving the quadratic equation for u results in the two solutions:

$u = 2 \pm 2i\,.$

Thus

$x^3 = 2 \pm 2i\,.$

Concentrating on finding the three cube roots of 2 + 2i – the other three solutions for x (the three cube roots of 2 - 2i ) will be their complex conjugates – rewriting the right-hand side using Euler's formula:

$x^3 = 2^{\tfrac{3}{2}}e^{\tfrac{1}{4}\pi i} = 2^{\tfrac{3}{2}}e^{\tfrac{8k+1}{4}\pi i}\,$

(since e^2kπi = 1), gives the three solutions:

$x = 2^{\tfrac{1}{2}}e^{\tfrac{8k+1}{12}\pi i}\,,~k = 0, 1, 2\,.$

Using Eulers' formula again together with trigonometric identities such as cos(π/12) = (√2 + √6) / 4, and adding the complex conjugates, gives the complete collection of solutions as:

$x_{1,2} = -1 \pm i,\,$

$x_{3,4} = \frac{1 + \sqrt{3}}{2} \pm \frac{1 - \sqrt{3}}{2}i\,$

and

$x_{5,6} = \frac{1 - \sqrt{3}}{2} \pm \frac{1 + \sqrt{3}}{2}i.\,$

Derivations of the quadratic formula

By completing the square

The quadratic formula can be derived by the method of completing the square,^[16] so as to make use of the algebraic identity:

$x^2+2xh+h^2 = (x+h)^2.\,\!$

Dividing the quadratic equation

$ax^2+bx+c=0 \,\!$

by a (which is allowed because a is non-zero), gives:

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

or

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

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

$x^2+\frac{b}{a}x+\left( \frac{1}{2}\frac{b}{a} \right)^2 =-\frac{c}{a}+\left( \frac{1}{2}\frac{b}{a} \right)^2,\!$

which produces

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

The right side can be written as a single fraction, with common denominator 4a². This gives

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

Taking the square root of both sides yields

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

Isolating x, gives

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

By shifting ax²

ax² with vertex shifted from the origin to (x_V, y_V). a=-1 in this example.

The quadratic formula can be derived by starting with equation

$a(x-x_V)^2 + y_V = 0 \,\!$

which describes the parabola as ax² with the vertex shifted from the origin to (x_V, y_V).

Solving this equation for x is straightforward and results in

$x = x_V\pm\sqrt{-\frac{y_V}{a}}.$

Using Vieta's formulas for the vertex coordinates

$\begin{align} x_V &= \frac{-b}{2a}\\ y_V &= -\frac{b^2-4ac}{4a},\\ \end{align}$

the values of x can be written as

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

Note. The formulas for x_V and y_V can be derived by comparing the coefficients in

$ax^2+bx+c=0 \,\!$

and

$a(x-x_V)^2 + y_V = 0 \,\!.$

Rewriting the latter equation as

$ax^2 + (-2ax_V)x + (a{x_V}^2 + y_V) = 0$

and comparing with the former results in

$b=-2ax_V \!$

$c=a{x_V}^2 + y_V \!,$

from which Vieta's expressions for x_V and y_V can be derived.

By Lagrange resolvents

For more details on this topic, see Lagrange resolvents.

An alternative way of deriving the quadratic formula is via the method of Lagrange resolvents, which is an early part of Galois theory.^[17] A benefit of this method is that it generalizes to give the solution of cubic polynomials and quartic polynomials, and leads to Galois theory, which allows one to understand the solution of polynomials of any degree in terms of the symmetry group of their roots, the Galois group.

This approach focuses on the roots more than on rearranging the original equation. Given a monic quadratic polynomial

$x^2+px+q,\!$

assume that it factors as

$x^2+px+q=(x-\alpha)(x-\beta).\!$

Expanding yields

$x^2+px+q=x^2-(\alpha+\beta)x+\alpha \beta,\!$

where

$p=-(\alpha+\beta)\!$

and

$q=\alpha \beta.\!$

Since the order of multiplication does not matter, one can switch α and β and the values of p and q will not change: one says that p and q are symmetric polynomials in α and β. In fact, they are the elementary symmetric polynomials – any symmetric polynomial in α and β can be expressed in terms of α + β and αβ. The Galois theory approach to analyzing and solving polynomials is: given the coefficients of a polynomial, which are symmetric functions in the roots, can one "break the symmetry" and recover the roots? Thus solving a polynomial of degree n is related to the ways of rearranging ("permuting") n terms, which is called the symmetric group on n letters, and denoted S_n. For the quadratic polynomial, the only way to rearrange two terms is to swap them ("transpose" them), and thus solving a quadratic polynomial is simple.

To find the roots α and β, consider their sum and difference:

$\begin{align} r_1 &= \alpha + \beta\\ r_2 &= \alpha - \beta.\\ \end{align}$

These are called the Lagrange resolvents of the polynomial; notice that these depend on the order of the roots, which is the key point. One can recover the roots from the resolvents by inverting the above equations:

$\begin{align} \alpha &= \textstyle{\frac{1}{2}}\left(r_1+r_2\right)\\ \beta &= \textstyle{\frac{1}{2}}\left(r_1-r_2\right).\\ \end{align}$

Thus, solving for the resolvents gives the original roots.

Formally, the resolvents are called the discrete Fourier transform (DFT) of order 2, and the transform can be expressed by the matrix $\left(\begin{smallmatrix}1 & 1\\ 1 & -1\end{smallmatrix}\right),$ with inverse matrix $\left(\begin{smallmatrix}1/2 & 1/2\\ 1/2 & -1/2\end{smallmatrix}\right).$ The transform matrix is also called the DFT matrix or Vandermonde matrix.

Now r₁ = α + β is a symmetric function in α and β, so it can be expressed in terms of p and q, and in fact r₁ = − p, as noted above. Contrariwise, r₂ = α − β is not symmetric, since switching α and β yields − r₂ = β − α (formally, this is termed a group action of the symmetric group of the roots). Since r₂ is not symmetric, it cannot be expressed in terms of the polynomials p and q, as these are symmetric in the roots and thus so is any polynomial expression involving them. However, changing the order of the roots only changes r₂ by a factor of − 1, and thus the square $\scriptstyle r_2^2 = (\alpha - \beta)^2$ is symmetric in the roots, and thus expressible in terms of p and q. Using the equation

$(\alpha - \beta)^2 = (\alpha + \beta)^2 - 4\alpha\beta\!$

yields

$r_2^2 = p^2 - 4q\!$

and thus

$r_2 = \pm \sqrt{p^2 - 4q}\!$.

If one takes the positive root, breaking symmetry, one obtains:

$\begin{align} r_1 &= -p\\ r_2 &= \sqrt{p^2 - 4q}\\ \end{align}$

and thus

$\begin{align} \alpha &= \textstyle{\frac{1}{2}}\left(-p+\sqrt{p^2 - 4q}\right)\\ \beta &= \textstyle{\frac{1}{2}}\left(-p-\sqrt{p^2 - 4q}\right)\\ \end{align}$

Thus the roots are

$\textstyle{\frac{1}{2}}\left(-p \pm \sqrt{p^2 - 4q}\right)$

which is the quadratic formula. Substituting $\scriptstyle p=\tfrac{b}{a}, q=\tfrac{c}{a}\!$ yields the usual form for when a quadratic is not monic. The resolvents can be recognized as $\scriptstyle \frac{r_1}{2} = \frac{-p}{2}=\frac{-b}{2a}\!$ being the vertex, and $\scriptstyle r_2^2=p^2-4q\!$ is the discriminant (of a monic polynomial).

A similar but more complicated method works for cubic equations, where one has three resolvents and a quadratic equation (the "resolving polynomial") relating r₂ and r₃, which one can solve by the quadratic equation, and similarly for a quartic (degree 4) equation, whose resolving polynomial is a cubic, which can in turn be solved. However, the same method for a quintic equation yields a polynomial of degree 24, which does not simplify the problem, and in fact solutions to quintic equations in general cannot be expressed using only roots.

Other methods of root calculation

Alternative quadratic formula

In some situations it is preferable to express the roots in an alternate form.

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

This alternative requires c to be nonzero; for, if c is zero, the formula correctly gives zero as one root, but fails to give any second, non-zero root. Instead, one of the two choices for ∓ produces the indeterminate form 0/0, which is undefined. However, the alternative form works when a is zero (giving the unique solution as one root and division by zero again for the other), which the normal form does not (instead producing division by zero both times).

The roots are the same regardless of which expression we use; the alternate form is merely an algebraic variation of the common form:

$\begin{align} \frac{-b + \sqrt {b^2-4ac\ }}{2a} &{}= \frac{-b + \sqrt {b^2-4ac\ }}{2a} \cdot \frac{-b - \sqrt {b^2-4ac\ }}{-b - \sqrt {b^2-4ac\ }} \\ &{}= \frac{b^2 - (b^2 - 4ac)}{2a \left ( -b - \sqrt {b^2-4ac} \right ) } \\ &{}= \frac{4ac}{2a \left ( -b - \sqrt {b^2-4ac} \right ) } \\ &{}=\frac{2c}{-b - \sqrt {b^2-4ac\ }}. \end{align}$

The alternative formula can reduce loss of precision in the numerical evaluation of the roots, which may be a problem if one of the roots is much smaller than the other in absolute magnitude. In this case, b is very close to $\scriptstyle \pm\sqrt{x}$, and the subtraction in the numerator causes loss of significance.

A mixed approach avoids both all cancellation problems (only numbers of the same sign are added), and the problem of c being zero:

$\begin{align} x_1 &= \frac{-b - \sgn (b) \,\sqrt {b^2-4ac}}{2a}, \\ x_2 &= \frac{2c}{-b - \sgn (b) \,\sqrt {b^2-4ac}} = \frac{c}{ax_1}. \end{align}$

Here sgn denotes the sign function.

Floating point implementation

A careful floating point computer implementation differs a little from both forms to produce a robust result. Assuming the discriminant, b² − 4ac, is positive and b is nonzero, the code will be something like the following:^[18]

$q = -\tfrac12 \left( b + \sgn(b) \sqrt{b^2-4ac} \right) \,\!$

$x_1 = q/a \,\!$

$x_2 = c/q \,\!$

Here sgn(b) is the sign function, where sgn(b) is 1 if b is positive and −1 if b is negative; its use ensures that the quantities added are of the same sign, avoiding catastrophic cancellation. The computation of x₂ uses the fact that the product of the roots is c/a.

Vieta's formulas

Main article: Vieta's formulas

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

$x_1 + x_2 = -\frac{b}{a}$

and

$x_1 \ x_2 = \frac{c}{a}.$

These results follow immediately from the relation:

$\left( x - x_1 \right) \ \left( x-x_2 \right ) = x^2 \ - \left( x_1+x_2 \right)x +x_1 \ x_2 \ = 0 \ ,$

which can be compared term by term with:

$x^2 + (b/a)x +c/a = 0 \ .$

The first formula above yields a convenient expression when graphing a quadratic function. Since the graph is symmetric with respect to a vertical line through the vertex, when there are two real roots the vertex’s x-coordinate is located at the average of the roots (or intercepts). Thus the x-coordinate of the vertex is given by the expression:

$x_V = \frac {x_1 + x_2} {2} = -\frac{b}{2a}.$

The y-coordinate can be obtained by substituting the above result into the given quadratic equation, giving

$y_V = - \frac{b^2}{4a} + c = - \frac{ b^2 - 4ac} {4a}.$

Graph of two evaluations of the smallest root of a quadratic: direct evaluation using the quadratic formula (accurate at smaller b) and an approximation for widely spaced roots (accurate for larger b). The difference reaches a minimum at the large dots, and rounding causes squiggles in the curves beyond this minimum.

As a practical matter, Vieta's formulas provide a useful method for finding the roots of a quadratic in the case where one root is much smaller than the other. If |x ₂| << |x ₁|, then x ₁ + x ₂ ≈ x ₁, and we have the estimate:

$x_1 \approx -\frac{b}{a} \ .$

The second Vieta's formula then provides:

$x_2 = \frac{c}{a \ x_1} \approx -\frac{c}{b} \ .$

These formulas are much easier to evaluate than the quadratic formula under the condition of one large and one small root, because the quadratic formula evaluates the small root as the difference of two very nearly equal numbers (the case of large b), which causes round-off error in a numerical evaluation. The figure shows the difference between (i) a direct evaluation using the quadratic formula (accurate when the roots are near each other in value) and (ii) an evaluation based upon the above approximation of Vieta's formulas (accurate when the roots are widely spaced). As the linear coefficient b increases, initially the quadratic formula is accurate, and the approximate formula improves in accuracy, leading to a smaller difference between the methods as b increases. However, at some point the quadratic formula begins to lose accuracy because of round off error, while the approximate method continues to improve. Consequently the difference between the methods begins to increase as the quadratic formula becomes worse and worse.

This situation arises commonly in amplifier design, where widely separated roots are desired to insure a stable operation (see step response).

Trigonometric solution for complex roots

In the case of complex roots the roots can also be found trigonometrically.^[19]

Generalization of quadratic equation

The formula and its derivation 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

$\pm \sqrt {b^2-4ac}$

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. Note that even if a field does not contain a square root of some number, there is always a quadratic extension field which does, so the quadratic formula will always make sense as a formula in that extension field.

Characteristic 2

In a field of characteristic 2, the quadratic formula, which relies on 2 being a unit, does not hold. Consider the monic quadratic polynomial

$\displaystyle x^{2} + bx + c$

over a field of characteristic 2. If b = 0, then the solution reduces to extracting a square root, so the solution is

$\displaystyle x = \sqrt{c}$

and note that there is only one root since

$\displaystyle -\sqrt{c} = -\sqrt{c} + 2\sqrt{c} = \sqrt{c}.$

In summary,

$\displaystyle x^{2} + c = (x + \sqrt{c})^{2}.$

See quadratic residue for more information about extracting square roots in finite fields.

In the case that b ≠ 0, there are two distinct roots, but if the polynomial is irreducible, they cannot be expressed in terms of square roots of numbers in the coefficient field. Instead, define the 2-root R(c) of c to be a root of the polynomial x² + x + c, an element of the splitting field of that polynomial. One verifies that R(c) + 1 is also a root. In terms of the 2-root operation, the two roots of the (non-monic) quadratic ax² + bx + c are

$\frac{b}{a}R\left(\frac{ac}{b^2}\right)$

and

$\frac{b}{a}\left(R\left(\frac{ac}{b^2}\right)+1\right).$

For example, let a denote a multiplicative generator of the group of units of F₄, the Galois field of order four (thus a and a + 1 are roots of x² + x + 1 over F₄). Because (a + 1)² = a, a + 1 is the unique solution of the quadratic equation x² + a = 0. On the other hand, the polynomial x² + ax + 1 is irreducible over F₄, but it splits over F₁₆, where it has the two roots ab and ab + a, where b is a root of x² + x + a in F₁₆.

This is a special case of Artin-Schreier theory.

See also

• Chakravala method
• Completing the square
• Cubic function
• Fundamental theorem of algebra
• Linear equation
• Parabola
• Periodic points of complex quadratic mappings
• Quadratic function
• Quadratic polynomial
• Quartic function
• Quintic function
• Solving quadratic equations with continued fractions

Notes

1. ^ Stillwell, p. 86.
2. ^ ^{a b} Stillwell, p. 87.
3. ^ [1] Jöran Friberg, A Geometric Algorithm with Solutions to Quadratic Equations in a Sumerian Juridical Document from Ur III Umma, CDLI, 2009.
4. ^ ^{a b c} BBC - h2g2 - The History Behind The Quadratic Formula
5. ^ David Eugene Smith (1958). "History of mathematics". Courier Dover Publications. p.134. ISBN 0486204294
6. ^ Katz, Victor J.; Barton, Bill (October 2007), "Stages in the History of Algebra with Implications for Teaching", Educational Studies in Mathematics (Springer Netherlands) 66 (2): 185–201 [190–1], doi:10.1007/s10649-006-9023-7 
7. ^ (Boyer 1991, "The Arabic Hegemony" p. 230) "Al-Khwarizmi here calls attention to the fact that what we designate as the discriminant must be positive: "You ought to understand also that when you take the half of the roots in this form of equation and then multiply the half by itself; if that which proceeds or results from the multiplication is less than the units above mentioned as accompanying the square, you have an equation." [...] Once more the steps in completing the square are meticulously indicated, without justification,"
8. ^ (Boyer 1991, "The Arabic Hegemony" p. 234) "The Algebra of al-Khwarizmi usually is regarded as the first work on the subject, but a recent publication in Turkey raises some questions about this. A manuscript of a work by 'Abd-al-Hamid ibn-Turk, entitled "Logical Necessities in Mixed Equations," was part of a book on Al-jabr wa'l muqabalah which was evidently very much the same as that by al-Khwarizmi and was published at about the same time - possibly even earlier. The surviving chapters on "Logical Necessities" give precisely the same type of geometric demonstration as al-Khwarizmi's Algebra and in one case the same illustrative example x² + 21 = 10x. In one respect 'Abd-al-Hamad's exposition is more thorough than that of al-Khwarizmi for he gives geometric figures to prove that if the discriminant is negative, a quadratic equation has no solution. Similarities in the works of the two men and the systematic organization found in them seem to indicate that algebra in their day was not so recent a development as has usually been assumed. When textbooks with a conventional and well-ordered exposition appear simultaneously, a subject is likely to be considerably beyond the formative stage. [...] Note the omission of Diophantus and Pappus, authors who evidently were not at first known in Arabia, although the Diophantine Arithmetica became familiar before the end of the tenth century."
9. ^ Katz, Victor J.; Barton, Bill (October 2007), "Stages in the History of Algebra with Implications for Teaching", Educational Studies in Mathematics (Springer Netherlands) 66 (2): 185–201 [191], doi:10.1007/s10649-006-9023-7 
10. ^ O'Connor, John J.; Robertson, Edmund F., "Arabic mathematics: forgotten brilliance?", MacTutor History of Mathematics archive, University of St Andrews, http://www-history.mcs.st-andrews.ac.uk/HistTopics/Arabic_mathematics.html . "Algebra was a unifying theory which allowed rational numbers, irrational numbers, geometrical magnitudes, etc., to all be treated as "algebraic objects"."
11. ^ Jacques Sesiano, "Islamic mathematics", p. 148, in Selin, Helaine; D'Ambrosio, Ubiratan (2000), Mathematics Across Cultures: The History of Non-Western Mathematics, Springer, ISBN 1402002602 
12. ^ The Equation that Couldn't be Solved
13. ^ Katz, Victor J.; Barton, Bill (October 2007), "Stages in the History of Algebra with Implications for Teaching", Educational Studies in Mathematics (Springer Netherlands) 66 (2): 185–201 [193], doi:10.1007/s10649-006-9023-7 
14. ^ Struik, D. J.; Stevin, Simon (1958), The Principal Works of Simon Stevin, II-B, C. V. Swets & Zeitlinger, p. 470, http://www.historyofscience.nl/works_detail.cfm?RecordId=2702 , Extract page 470
15. ^ Heaton, H. (1896) A Method of Solving Quadratic Equations, American Mathematical Monthly 3(10), 236–237.
16. ^ Rich, Barnett; Schmidt, Philip (2004), Schaum's Outline of Theory and Problems of ELEMENTARY ALGEBRA, The McGraw-Hill Companies, ISBN 0-07-141083-X, http://books.google.com/?id=8PRU9cTKprsC , Chapter 13 §4.4, p. 291
17. ^ Prasolov, Viktor; Solovyev, Yuri (1997), Elliptic functions and elliptic integrals, AMS Bookstore, ISBN 978 0 82180587 9, http://books.google.com/?id=fcp9IiZd3tQC , §6.2, p. 134
18. ^ Press, William H.; Flannery, Brian P.; Teukolsky, Saul A.; Vetterling, William T. (1992), Numerical Recipes in C (Second ed.), http://www.nrbook.com/a/bookcpdf.php , Section 5.6: "Quadratic and Cubic Equations.
19. ^ Simons, Stuart, "Alternative approach to complex roots of real quadratic equations", Mathematical Gazette 93, March 2009, 91-92.

References

• Al-Dīnawarī, Abū Ḥanīfa. 820. Al-Kitāb al-mukhtaṣar fī hīsāb al-ğabr wa’l-muqābala.
• Stillwell, John, Mathematics and Its History, Springer; 2nd edition (January 27, 2004). ISBN 0-387-95336-1.

External links

• Weisstein, Eric W., "Quadratic equations" from MathWorld.
• 101 uses of a quadratic equation
• 101 uses of a quadratic equation: Part II
• Step-by-step instructions on using the quadratic formula for any input