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A quadratic function with roots x=1 and x=4

In elementary algebra, the quadratic formula is the solution of the quadratic equation. There are other ways to solve the quadratic equation instead of using the quadratic formula, such as factoring, completing the square, or graphing. Using the quadratic formula is often the most convenient way.

${\displaystyle ax^{2}+bx+c=0.}$

Here x represents an unknown, while a, b, and c are constants with a not equal to 0. One can verify that the quadratic formula satisfies the quadratic equation by inserting the former into the latter. With the above parameterization, the quadratic formula is:

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

Each of the solutions given by the quadratic formula is called a root of the quadratic equation. Geometrically, these roots represent the x values at which any parabola, explicitly given as y = axBags Bags Ppge Silver Transparent Chain Handle Women Messenger Blue Wild Shoulder Bag Gelatin 2 + bx + c, crosses the x-axis. As well as being a formula that will yield the zeros of any parabola, the quadratic formula will give the axis of symmetry of the parabola, and it can be used to immediately determine how many real zeros the quadratic equation has.

## Derivation of the formula

The quadratic formula can be derived with a simple application of technique of completing the square.[1][2] For this reason, the derivation is sometimes left as an exercise for students, who can thus experience rediscovery of this important formula.[3][4] The explicit derivation is as follows.

Divide the quadratic equation by a, which is allowed because a is non-zero:

${\displaystyle x^{2}+{\frac {b}{a}}x+{\frac {c}{a}}=0.}$
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Subtract c/a from both sides of the equation, yielding:

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

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.

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

which produces:

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

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

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

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

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

Isolating x gives the quadratic formula:

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

The plus-minus symbol "±" indicates that both

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

are solutions of the quadratic equation.[5] There are many alternatives of this derivation with minor differences, mostly concerning the manipulation of a.

Some sources, particularly older ones, use alternative parameterizations of the quadratic equation such as ax2 − 2bx + c = 0[6] or ax2 + 2bx + c = 0,[7] where b 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:

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

## Geometrical significance

Graph of y = ax2 + bx + c, where a and the discriminant b2 − 4ac are positive, with
• Roots and y-intercept in red
• Vertex and axis of symmetry in blue
• Focus and directrix in pink
Visualisation of the complex roots of y = ax2 + bx + c: the parabola is rotated 180° about its vertex ( orange). Its x-intercepts are rotated 90° around their mid-point, and the Cartesian plane is interpreted as the complex plane ( green). [8]

In terms of coordinate geometry, a parabola is a curve whose (x, y)-coordinates are described by a second-degree polynomial, i.e. any equation of the form:

${\displaystyle y=p(x)=a_{2}x^{2}+a_{1}x+a_{0},}$

where p represents the polynomial of degree 2 and a0, a1, and a2 ≠ 0 are constant coefficients whose subscripts correspond to their respective term's degree. The geometrical interpretation of the quadratic formula is that it defines the points on the x-axis where the parabola will cross the axis. Additionally, if the quadratic formula was looked at as two terms,

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

the axis of symmetry appears as the line x = −b/2a. The other term, b2 − 4ac/2a, gives the distance the zeros are away from the axis of symmetry, where the plus sign represents the distance to the right, and the minus sign represents the distance to the left.

If this distance term were to decrease to zero, the value of the axis of symmetry would be the x value of the only zero, that is, there is only one possible solution to the quadratic equation. Algebraically, this means that b2 − 4ac = 0, or simply b2 − 4ac = 0 (where the left-hand side is referred to as the discriminant). This is one of three cases, where the discriminant indicates how many zeros the parabola will have. If the discriminant is positive, the distance would be non-zero, and there will be two solutions. However, there is also the case where the discriminant is less than zero, and this indicates the distance will be imaginary – or some multiple of the complex unit i, where i = −1 – and the parabola's zeros will be complex numbers. The complex roots will be complex conjugates, where the real part of the complex roots will be the value of the axis of symmetry. There will be no real values of x where the parabola crosses the x-axis.

## Historical development

The earliest methods for solving quadratic equations were geometric. Babylonian cuneiform tablets contain problems reducible to solving quadratic equations.[9] The Egyptian Berlin Papyrus, dating back to the Middle Kingdom (2050 BC to 1650 BC), contains the solution to a two-term quadratic equation.[10]

The Greek mathematician Euclid (circa 300 BC) used geometric methods to solve quadratic equations in Book 2 of his Elements, an influential mathematical treatise.Cloth Deep Shirtee xt003 Helles 38cm Blue Womens Blue 42cm 38cm Sdzthy5r Grün 42cm Bag 5743 Cotton w7wHY Rules for quadratic equations appear in the Chinese The Nine Chapters on the Mathematical Art circa 200 BC.[12][13] In his work Arithmetica, the Greek mathematician Diophantus (circa 250 BC) solved quadratic equations with a method more recognizably algebraic than the geometric algebra of Euclid.Cloth Deep Shirtee xt003 Helles 38cm Blue Womens Blue 42cm 38cm Sdzthy5r Grün 42cm Bag 5743 Cotton w7wHY His solution gives only one root, even when both roots are positive.[14]

The Indian mathematician Brahmagupta (597–668 AD) explicitly described the quadratic formula in his treatise Brāhmasphuṭasiddhānta published in 628 AD,[15] but written in words instead of symbols.[16] His solution of the quadratic equation ax2 + bx = c was 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."[17] This is equivalent to:

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

The 9th-century Persian mathematician al-Khwārizmī, influenced by earlier Greek and Indian mathematicians, solved quadratic equations algebraically.[18] The quadratic formula covering all cases was first obtained by Simon Stevin in 1594.[19] In 1637 René Descartes published La Géométrie containing special cases of the quadratic formula in the form we know today.[citation needed] The first appearance of the general solution in the modern mathematical literature appeared in an 1896 paper by Henry Heaton.[20]

## Other derivations

Many alternative derivations of the quadratic formula are in the literature. These derivations may be simpler than the standard completing the square method, may represent interesting applications of other algebraic techniques, or may offer insight into other areas of mathematics.

### Alternate method of completing the square

The majority of algebra texts published over the last several decades teach Inclined New Gray Ocio Single Bolsas Layers Gris Inclinados Nueva Moda Summer Solo Bags Tres Bolsos Three Fashion De Handbags Sjmmbb Sjmmbb 26x14x18cm 26x14x18cm Capas Hombro De Verano De gris Leisure Gray Shoulder Twpgfgq0: (1) divide each side by a to make the equation monic, (2) rearrange, (3) then add (bWild Bag Women Bags Ppge Blue Silver Transparent Shoulder Messenger Handle Bags Chain Gelatin /2a)2 to both sides to complete the square.

As pointed out by Larry Hoehn in 1975, completing the square can be accomplished by a different sequence that leads to a simpler sequence of intermediate terms: (1) multiply each side by 4a, (2) rearrange, (3) then add b2.[21]

In other words, the quadratic formula can be derived as follows: