Quadratic equation

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

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

where a ≠ 0.

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

Quadratic equations are called quadratic because quadratus is Latin for "square"; in the leading term the variable is squared.

The graph of a real valued quadratic function f(x) = ax2 + bx + c, of a real variable x, is a parabola.

The graph of a real valued quadratic function f(x) = ax² + bx + c, of a real variable x, is a parabola.

1 Quadratic formula
2 Discriminant
3 Geometry
4 Quadratic factorization
5 Application to higher-degree equations
6 History
7 Derivation
8 Alternative formula
9 Viète's formulas
10 Generalizations
- 10.1 Characteristic 2
11 See also
12 External links

[edit] Quadratic formula

A quadratic equation with real (or complex) coefficients has two (but not necessarily distinct) solutions, called roots, which may be real or complex, given by the quadratic formula:

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

where the symbol "±" indicates that both

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

and

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

are solutions.

[edit] Discriminant

In the above formula, the term underneath the square root sign:

$b^2 - 4ac,\,\!$

is called the discriminant of the quadratic equation.

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

If the discriminant is positive, there are two distinct roots, both of which are real numbers. 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, there is exactly one root, and that root is a real number. Sometimes called a double root, its value is:

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

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

$x_+ = \frac{-b}{2a} + i \left ( \frac{\sqrt {4ac - b^2}}{2a} \right )$

and

$x_- = \frac{-b}{2a} - i \left ( \frac{\sqrt {4ac - b^2}}{2a} \right ),$

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.

[edit] 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 touches the x-axis, x = −1 and x = 2, are the roots of the quadratic equation: x2 − x − 2 = 0.

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 touches the x-axis, x = −1 and x = 2, are the roots of the quadratic equation: x² − x − 2 = 0.

The roots of the quadratic equation

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

are also the zeros 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 zeros 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.

[edit] Quadratic factorization

The term x − r is a factor of the polynomial

$x^2+bx+c,\,\!$

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

$x^2+bx+c=0.\,\!$

It follows from the quadratic formula that

$x^2+bx+c = (x - \frac{-b + \sqrt {b^2-4c}}{2}) (x - \frac{-b - \sqrt {b^2-4c}}{2}).\,\!$

Or equivalently that:

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

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

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

[edit] Application to higher-degree equations

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

$2x^6 + 3x^3 + 5 = 0,\,$

which can be written as

$2u^2 + 3u + 5 = 0, \$

where

$u = x^3 \$ .

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, the quadratic formula, or completing the square.

Generally speaking, if the polynomial is quadratic in some variable u where

$u = x^n \,\;$

then the quadratic equation can be used to help find solutions.

[edit] History

On clay tablets dated between 1800 BC and 1600 BC, the ancient Babylonians left the earliest evidence of the discovery of quadratic equations, and also gave early methods for solving them. Indian mathematician Baudhayana who wrote a Sulba Sutra in ancient India circa 8th century BC first used quadratic equations of the form ax² = c and ax² + bx = c and also gave methods for solving them.

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. Euclid produced a more abstract geometrical method around 300 BC. The Bakshali Manuscript written in India between 200 BC and 400 CE introduced the general algebraic formula for solving quadratic equations, and also introduced quadratic indeterminate equations (origin of type ax/c = y).

The first mathematician to have found negative solutions with the general algebraic formula, was Brahmagupta (India, 7th century). Muḥammad ibn Mūsā al-Ḵwārizmī (Persia, 9th century) developed a set of formulae that worked for positive solutions. Abraham bar Hiyya Ha-Nasi (also known by the Latin name Savasorda) introduced the complete solution to Europe in his book Liber embadorum in the 12th century. Bhaskara II (India, 12th century) solved quadratic equations with more than one unknown.

Shridhara (India, 9th century) was one of the first mathematicians to give a general rule for solving a quadratic equation. His original work is lost but Bhaskara II later quotes Shridhara's rule:

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. [1]

[edit] Derivation

The quadratic formula is derived by the method of completing the square, so as to make use of the algebraic identity:

$x^2+2xy+y^2 = (x+y)^2.\,\!$

Dividing the quadratic equation

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

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

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

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

The quadratic equation is now in a form in which the method of completing the square can be applied. To "complete the square" is to find some constant k such that

$x^2 + \frac{b}{a} x +k = x^2+2xy+y^2,\,\!$

for some y. But for the above equation to hold it must be the case that

$y = \frac{b}{2a}\,\!$

and

$k = y^2,\,\!$

thus

$k = \frac{b^2}{4a^2}.\,\!$

Adding this constant to both sides of the equation

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

produces

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

The left side is now a perfect square; it is the square of

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

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

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

Isolating x, gives

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

[edit] Alternative formula

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

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

However, it imposes the additional requirement that c be nonzero. If c is zero, this formula correctly gives zero as one root, but fails to give any second, non-zero root. (When c is zero we have division of zero by zero, which is undefined.)

The actual values of the roots must be the same regardless of which expression we use, so the alternate form is merely an algebraic variation of the common form. For example,

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

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.

$t := -(b + \sgn(b) \sqrt{b^2-4ac})/2 \,\!$

$r_{1} := t/a \,\!$

$r_{2} := c/t \,\!$

Here sgn(b) is the sign function, giving +1 if b is positive and −1 if b is negative; its use ensures that we always add two quantities of the same sign, avoiding catastrophic cancellation. The computation of r₂ uses the fact that the product of the roots is c/a.

[edit] Viète's formulas

Viète'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:

$x_+ + x_- = -\frac{b}{a}$

and

$x_+ \cdot x_- = \frac{c}{a}.$

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:

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

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

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

[edit] Generalizations

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 $b 2 - 4 a c,$ 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.

[edit] 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 x² + bx + c over a field of characteristic 2. If b = 0, then the solution reduces to extracting a square root, so the solution is x = √c and note that there is only one root since –√c = –√c + 2√c = √c. In summary, x² + c = (x + √c)². Confer 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 x + ax + 1 is irreducible over F₄, but 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

[edit] See also

[edit] External links

Weisstein, Eric W., Quadratic equations at MathWorld.
Quadratic equation solver, plus solvers for cubic and quartic equations
Quadratic equation solver
Solve quadratic equations, see work shown and draw graphs
101 uses of a quadratic equation part I Part II
Quadratic graphical explorer Interactive applet. Sliders for a,b,c show effects on a graph.

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Categories: Elementary algebra | Equations