Quadratic equation

An alternative way of deriving the quadratic formula is via the method of Lagrange resolvents, which is an early part of Galois theory.^[25] 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

${\displaystyle x^{2}+px+q,\!}$

assume that it factors as

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

Expanding yields

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

where

${\displaystyle p=-(\alpha +\beta )\!}$

and

${\displaystyle q=\alpha \beta .\!}$

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 ${\displaystyle S_{n}.}$ 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:

${\displaystyle {\begin{aligned}r_{1}&=\alpha +\beta \\r_{2}&=\alpha -\beta .\\\end{aligned}}}$

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:

${\displaystyle {\begin{aligned}\alpha &=\textstyle {\frac {1}{2}}\left(r_{1}+r_{2}\right)\\\beta &=\textstyle {\frac {1}{2}}\left(r_{1}-r_{2}\right).\\\end{aligned}}}$

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 ${\displaystyle \left({\begin{smallmatrix}1&1\\1&-1\end{smallmatrix}}\right),}$ with inverse matrix ${\displaystyle \left({\begin{smallmatrix}1/2&1/2\\1/2&-1/2\end{smallmatrix}}\right).}$ The transform matrix is also called the DFT matrix or Vandermonde matrix.

Now ${\displaystyle r_{1}=\alpha +\beta }$ is a symmetric function in α and β, so it can be expressed in terms of p and q, and in fact ${\displaystyle r_{1}=-p,}$ as noted above. But ${\displaystyle r_{2}=\alpha -\beta }$ is not symmetric, since switching α and β yields ${\displaystyle -r_{2}=\beta -\alpha }$ (formally, this is termed a group action of the symmetric group of the roots). Since ${\displaystyle r_{2}}$ 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 ${\displaystyle r_{2}}$ by a factor of ${\displaystyle -1,}$ and thus the square ${\displaystyle \scriptstyle r_{2}^{2}=(\alpha -\beta )^{2}}$ is symmetric in the roots, and thus expressible in terms of p and q. Using the equation

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

yields

${\displaystyle r_{2}^{2}=p^{2}-4q\!}$

and thus

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

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

${\displaystyle {\begin{aligned}r_{1}&=-p\\r_{2}&={\sqrt {p^{2}-4q}}\\\end{aligned}}}$

and thus

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

Thus the roots are

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

which is the quadratic formula. Substituting ${\displaystyle \scriptstyle p={\tfrac {b}{a}},q={\tfrac {c}{a}}\!}$ yields the usual form for when a quadratic is not monic. The resolvents can be recognized as ${\displaystyle \scriptstyle {\frac {r_{1}}{2}}={\frac {-p}{2}}={\frac {-b}{2a}}\!}$ being the vertex, and ${\displaystyle \scriptstyle r_{2}^{2}=p^{2}-4q\!}$ 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 ${\displaystyle r_{2}}$ and ${\displaystyle r_{3},}$ 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.

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

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

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 alternative form is merely an algebraic variation of the common form:

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

A careful floating point computer implementation combines several strategies to produce a robust result. Assuming the discriminant, b² − 4ac, is positive and b is nonzero, the code will be something like the following:^[26]

${\displaystyle q=-{\tfrac {1}{2}}\left(b+\operatorname {sgn}(b){\sqrt {b^{2}-4ac}}\right)\,\!}$

${\displaystyle x_{1}=q/a\,\!}$

${\displaystyle x_{2}=c/q\,\!}$

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. Note that while the above formulation avoids catastrophic cancellation between b and ${\displaystyle {\sqrt {b^{2}-4ac}}}$, there remains a form of cancellation between the terms b² and −4ac of the discriminant, which can still lead to loss of up to half of correct significant figures.^[5][11] The discriminant b²−4ac needs to be computed in arithmetic of twice the precision of the result to avoid this (e.g. quad precision if the final result is to be accurate to full double precision).^[27] This can be in the form of a fused multiply-add operation.^[5]

To illustrate this, consider the following quadratic equation, adapted from Kahan (2004):^[5]

94906265.625x² − 189812534x + 94906268.375

This equation has Δ = 7.5625 and has roots

x₁ = 1.000000028975958

x₂ = 1.000000000000000.

However, when computed using IEEE 754 double-precision arithmetic corresponding to 15 to 17 significant digits of accuracy, Δ is rounded to 0.0, and the computed roots are

x₁ = 1.000000014487979

x₂ = 1.000000014487979

which are both false after the eighth significant digit. This is despite the fact that superficially, the problem seems to require only eleven significant digits of accuracy for its solution.

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:

${\displaystyle x_{1}+x_{2}=-{\frac {b}{a}}}$

and

${\displaystyle x_{1}\ x_{2}={\frac {c}{a}}.}$

These results follow immediately from the relation:

${\displaystyle \left(x-x_{1}\right)\ \left(x-x_{2}\right)=x^{2}\ -\left(x_{1}+x_{2}\right)x+x_{1}\ x_{2}\ =0\ ,}$

which can be compared term by term with:

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

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:

${\displaystyle x_{V}={\frac {x_{1}+x_{2}}{2}}=-{\frac {b}{2a}}.}$

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

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

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:

${\displaystyle x_{1}\approx -{\frac {b}{a}}\ .}$

The second Vieta's formula then provides:

${\displaystyle x_{2}={\frac {c}{a\ x_{1}}}\approx -{\frac {c}{b}}\ .}$

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 ensure a stable operation (see step response).

In the days before calculators, people would use mathematical tables—lists of numbers showing the results of calculation with varying arguments—to simplify and speed up computation. Tables of logarithms and trigonometric functions were common in math and science textbooks. Specialized tables were published for applications such as astronomy, celestial navigation and statistics. Methods of numerical approximation existed, called prosthaphaeresis, that offered shortcuts around time-consuming operations such as multiplication and taking powers and roots.^[10] Astronomers, especially, were concerned with methods that could speed up the long series of computations involved in celestial mechanics calculations.

It is within this context that we may understand the development of means of solving quadratic equations by the aid of trigonometric substitution. Consider the following alternate form of the quadratic equation,

[1]   ${\displaystyle ax^{2}+bx\pm c=0,}$

where the sign of the ± symbol is chosen so that a and c may both be positive. By substituting

[2]   ${\displaystyle x={\sqrt {c/a}}\tan \theta }$

and then multiplying through by cos²θ, we obtain

[3]   ${\displaystyle \sin ^{2}\theta +{\frac {b}{\sqrt {ac}}}\sin \theta \cos \theta \pm \cos ^{2}\theta =0.}$

Introducing functions of 2θ and rearranging, we obtain

[4]   ${\displaystyle \tan 2\theta _{n}=+2{\frac {\sqrt {ac}}{b}},}$

[5]   ${\displaystyle \sin 2\theta _{p}=-2{\frac {\sqrt {ac}}{b}},}$

where the subscripts n and p correspond, respectively, to the use of a negative or positive sign in equation [1]. Substituting the two values of θ_n or θ_p found from equations [4] or [5] into [2] gives the required roots of [1]. Complex roots occur in the solution based on equation [5] if the absolute value of sin 2θ_p exceeds unity. The amount of effort involved in solving quadratic equations using this mixed trigonometric and logarithmic table look-up strategy was two-thirds the effort using logarithmic tables alone.^[28] Calculating complex roots would require using a different trigonometric form.^[29]

x₂ = −10^{(0.2192318 + 0.2706462)} = −3.08943

The quadratic equation may be solved geometrically in a number of ways. One way is via Lill's method. The three coefficients a, b, c are drawn with right angles between them as in SA, AB, and BC in the accompanying diagram. A circle is drawn with the start and end point SC as a diameter. If this cuts the middle line AB of the three then the equation has a solution, and the solutions are given by negative of the distance along this line from A divided by the first coefficient a or SA. If a is 1 the coefficients may be read off directly. Thus the solutions in the diagram are −AX1/SA and −AX2/SA.^[30]

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

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

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.

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

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

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

${\displaystyle \displaystyle x={\sqrt {c}}}$

and note that there is only one root since

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

In summary,

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

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

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

and

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

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.

• Boyer, Carl B.; Merzbach, Uta C. (1991), "The Arabic Hegemony", A History of Mathematics (2nd ed.), Wiley, ISBN 0-471-54397-7

• "Quadratic equation", Encyclopedia of Mathematics, EMS Press, 2001 [1994]
• Weisstein, Eric W. "Quadratic equations". 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