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.

The general quadratic equation is

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:

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* = *ax*^{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.

## Contents

## Derivations of the formula

### By using completing the square method

The quadratic formula can be derived with a simple application of technique of completing the square.^{[1]}^{[2]}^{[3]}^{[4]} The two derivations are as follows:

#### Method 1

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

Subtract *c*/*a* from both sides of the equation, yielding:

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.

which produces:

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

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

Isolating *x* gives the quadratic formula:

The plus-minus symbol "±" indicates that

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

The quadratic formula may also be written as:

which may be simplified to:

This version of the formula is convenient when complex roots are accepted. Then the expression outside the square root will be the real part and the square root expression will be the imaginary part. The expression inside the square root is a discriminant.

Some sources, particularly older ones, use alternative parameterizations of the quadratic equation such as *ax*^{2} − 2*bx* + *c* = 0^{[6]} or *ax*^{2} + 2*bx* + *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:

#### Method 2

The majority of algebra texts published over the last several decades teach completing the square using the sequence presented earlier: (1) divide each side by *a* to make the equation monic, (2) rearrange, (3) then add (*b*/2*a*)^{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 4*a*, (2) rearrange, (3) then add *b*^{2}.^{[8]}

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

This actually represents an ancient derivation of the quadratic formula and was known to the Hindus at least as far back as 1025.^{[9]} Compared with the derivation in standard usage, this alternate derivation is shorter, involves fewer computations with literal coefficients, avoids fractions until the last step, has simpler expressions, and uses simpler mathematics. As Hoehn states, "it is easier 'to add the square of *b*' than it is 'to add the square of half the coefficient of the *x* term'".^{[8]}

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

### By substitution

Another technique is solution by substitution.^{[10]} In this technique, we substitute *x* = *y* + *m* into the quadratic to get:

Expanding the result and then collecting the powers of *y* produces:

We have not yet imposed a second condition on *y* and *m*, so we now choose *m* so that the middle term vanishes. That is, 2*am* + *b* = 0 or *m* = −*b*/2*a*. Subtracting the constant term from both sides of the equation (to move it to the right hand side) and then dividing by *a* gives:

Substituting for *m* gives:

Therefore,

substituting *x* = *y* + *m* = *y* −*b*/2*a* provides the quadratic formula

### By using algebraic identities

The following method was used by many historical mathematicians:^{[11]}

Let the roots of the standard quadratic equation be *r*_{1} and *r*_{2}. The derivation starts by recalling the identity:

Taking the square root on both sides, we get:

Since the coefficient *a* ≠ 0, we can divide the standard equation by *a* to obtain a quadratic polynomial having the same roots. Namely,

From this we can see that the sum of the roots of the standard quadratic equation is given by −*b*/*a*, and the product of those roots is given by *c*/*a*.
Hence the identity can be rewritten as:

Now,

Since *r*_{2} = −*r*_{1} − *b*/*a*, if we take

then we obtain

and if we instead take

then we calculate that

Combining these results by using the standard shorthand ±, we have that the solutions of the quadratic equation are given by:

### By Lagrange resolvents

An alternative way of deriving the quadratic formula is via the method of Lagrange resolvents,^{[12]} which is an early part of Galois theory.^{[13]}
This method can be generalized to give the roots of cubic polynomials and quartic polynomials, and leads to Galois theory, which allows one to understand the solution of algebraic equations 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

assume that it factors as

Expanding yields

where *p* = −(*α* + *β*) and *q* = *αβ*.

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 can say 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 *��β* 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:

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

Thus, solving for the resolvents gives the original roots.

Now *r*_{1} = *α* + *β* is a symmetric function in *α* and *β*, so it can be expressed in terms of *p* and *q*, and in fact *r*_{1} = −*p* as noted above. But *r*_{2} = *α* − *β* is not symmetric, since switching *α* and *β* yields −*r*_{2} = *β* − *α* (formally, this is termed a group action of the symmetric group of the roots). Since *r*_{2} is not symmetric, it cannot be expressed in terms of the coefficients *p* and *q*, as these are symmetric in the roots and thus so is any polynomial expression involving them. Changing the order of the roots only changes *r*_{2} by a factor of −1, and thus the square *r*_{2}^{2} = (*α* − *β*)^{2} is symmetric in the roots, and thus expressible in terms of *p* and *q*. Using the equation

yields

and thus

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

and thus

Thus the roots are

which is the quadratic formula. Substituting *p* = *b*/*a*, *q* = *c*/*a* yields the usual form for when a quadratic is not monic. The resolvents can be recognized as *r*_{1}/2 = −*p*/2 = −*b*/2*a* being the vertex, and *r*_{2}^{2} = *p*^{2} − 4*q* 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 *r*_{2} and *r*_{3}, which one can solve by the quadratic equation, and similarly for a quartic equation (degree 4), whose resolving polynomial is a cubic, which can in turn be solved.^{[12]} 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.

### By extrema

Knowing the value of *x* in the functional extreme point makes it possible to solve only for the increase (or decrease) needed in *x* to solve the quadratic equation. This method first uses differentiation to find the *x* value at the extremum, called *x*_{ext}. We then solve for the value, *q*, that ensures that *f*(*x*_{ext} + *q*) = 0. While this may not be the most intuitive method, it ensures that the mathematics is straightforward.

Setting the above differential to zero will give us the extrema of the quadratic function

We define *q* as follows:

Here *x*_{0} is the value of *x* that solves the quadratic equation. The sum of *x*_{ext} and the variable of interest, *q*, is plugged in to the quadratic equation

The value of *x* in the extreme point is then added to both sides of the equation

This gives the quadratic formula. This way one avoids the technique of completing the square and much more complicated math is not needed. Note this solution is very similar to solving deriving the formula by substitution.

### By splitting into real and imaginary parts

Consider the equation

where is a complex number and where *a*, *b*, and *c* are real numbers. Then

This splits into two equations, the real part:

and the imaginary part:

Assuming that then divide the second equation by *y*:

and solve for *x*:

Substitute this value for *x* into the first equation and solve for *y*:

Since , then

Even though *y* was assumed to be non-zero, this last formula works for any roots of the original equation, whereas assuming that turns out to be of not much help (trivial and circular).

## Historical development

The earliest methods for solving quadratic equations were geometric. Babylonian cuneiform tablets contain problems reducible to solving quadratic equations.^{[15]} The Egyptian Berlin Papyrus, dating back to the Middle Kingdom (2050 BC to 1650 BC), contains the solution to a two-term quadratic equation.^{[16]}

The Greek mathematician Euclid (circa 300 BC) used geometric methods to solve quadratic equations in Book 2 of his *Elements*, an influential mathematical treatise.^{[17]} Rules for quadratic equations appear in the Chinese *The Nine Chapters on the Mathematical Art* circa 200 BC.^{[18]}^{[19]} 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.^{[17]} His solution gives only one root, even when both roots are positive.^{[20]}

The Indian mathematician Brahmagupta (597–668 AD) explicitly described the quadratic formula in his treatise *Brāhmasphuṭasiddhānta* published in 628 AD,^{[21]} but written in words instead of symbols.^{[22]} His solution of the quadratic equation *ax*^{2} + *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."^{[23]}
This is equivalent to:

The 9th-century Persian mathematician Muḥammad ibn Mūsā al-Khwārizmī solved quadratic equations algebraically.^{[24]} The quadratic formula covering all cases was first obtained by Simon Stevin in 1594.^{[25]} 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.^{[26]}

## Significant uses

### Geometrical significance

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:

where *p* represents the polynomial of degree 2 and *a*_{0}, *a*_{1}, and *a*_{2} ≠ 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,

the axis of symmetry appears as the line *x* = −*b*/2*a*. The other term, √*b*^{2} − 4*ac*/2*a*, 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 √*b*^{2} − 4*ac* = 0, or simply *b*^{2} − 4*ac* = 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.

### Dimensional analysis

If the constants *a*, *b*, and/or *c* are not unitless, then the units of *x* must be equal to the units of *b*/*a*, due to the requirement that *ax*^{2} and *bx* agree on their units. Furthermore, by the same logic, the units of *c* must be equal to the units of *b*^{2}/*a*, which can be verified without solving for *x*. This can be a powerful tool for verifying that a quadratic expression of physical quantities has been set up correctly, prior to solving it.

## See also

## References

**^**Rich, Barnett; Schmidt, Philip (2004),*Schaum's Outline of Theory and Problems of Elementary Algebra*, The McGraw–Hill Companies, ISBN 0-07-141083-X, Chapter 13 §4.4, p. 291**^**Li, Xuhui.*An Investigation of Secondary School Algebra Teachers' Mathematical Knowledge for Teaching Algebraic Equation Solving*, p. 56 (ProQuest, 2007): "The quadratic formula is the most general method for solving quadratic equations and is derived from another general method: completing the square."**^**Rockswold, Gary.*College algebra and trigonometry and precalculus*, p. 178 (Addison Wesley, 2002).**^**Beckenbach, Edwin et al.*Modern college algebra and trigonometry*, p. 81 (Wadsworth Pub. Co., 1986).**^**Sterling, Mary Jane (2010),*Algebra I For Dummies*, Wiley Publishing, p. 219, ISBN 978-0-470-55964-2**^**Kahan, Willian (November 20, 2004),*On the Cost of Floating-Point Computation Without Extra-Precise Arithmetic*(PDF), retrieved 2012-12-25**^**"Quadratic Formula",*Proof Wiki*, retrieved 2016-10-08- ^
^{a}^{b}Hoehn, Larry (1975). "A More Elegant Method of Deriving the Quadratic Formula".*The Mathematics Teacher*.**68**(5): 442–443. **^**Smith, David E. (1958).*History of Mathematics, Vol. II*. Dover Publications. p. 446. ISBN 0486204308.**^**Joseph J. Rotman. (2010). Advanced modern algebra (Vol. 114). American Mathematical Soc. Section 1.1**^**Debnath, L. (2009). The legacy of Leonhard Euler–a tricentennial tribute. International Journal of Mathematical Education in Science and Technology, 40(3), 353–388. Section 3.6- ^
^{a}^{b}Clark, A. (1984).*Elements of abstract algebra*. Courier Corporation. p. 146. **^**Prasolov, Viktor; Solovyev, Yuri (1997),*Elliptic functions and elliptic integrals*, AMS Bookstore, ISBN 978-0-8218-0587-9, §6.2, p. 134**^**"Complex Roots Made Visible – Math Fun Facts". Retrieved 1 October 2016.**^**Irving, Ron (2013).*Beyond the Quadratic Formula*. MAA. p. 34. ISBN 978-0-88385-783-0.**^***The Cambridge Ancient History Part 2 Early History of the Middle East*. Cambridge University Press. 1971. p. 530. ISBN 978-0-521-07791-0.- ^
^{a}^{b}Irving, Ron (2013).*Beyond the Quadratic Formula*. MAA. p. 39. ISBN 978-0-88385-783-0. **^**Aitken, Wayne. "A Chinese Classic: The Nine Chapters" (PDF). Mathematics Department, California State University. Retrieved 28 April 2013.**^**Smith, David Eugene (1958).*History of Mathematics*. Courier Dover Publications. p. 380. ISBN 978-0-486-20430-7.**^**Smith, David Eugene (1958).*History of Mathematics*. Courier Dover Publications. p. 134. ISBN 0-486-20429-4.**^**Bradley, Michael.*The Birth of Mathematics: Ancient Times to 1300*, p. 86 (Infobase Publishing 2006).**^**Mackenzie, Dana.*The Universe in Zero Words: The Story of Mathematics as Told through Equations*, p. 61 (Princeton University Press, 2012).**^**Stillwell, John (2004).*Mathematics and Its History (2nd ed.)*. Springer. p. 87. ISBN 0-387-95336-1.**^**Irving, Ron (2013).*Beyond the Quadratic Formula*. MAA. p. 42. ISBN 978-0-88385-783-0.**^**Struik, D. J.; Stevin, Simon (1958),*The Principal Works of Simon Stevin, Mathematics*(PDF),**II–B**, C. V. Swets & Zeitlinger, p. 470**^**Heaton, H. (1896)*A Method of Solving Quadratic Equations*, American Mathematical Monthly**3**(10), 236–237.