Quadratic formula

Not to be confused with quadratic function or quadratic equation.

The quadratic function y = 1/2x² − 5/2x + 2, with roots x = 1 and x = 4.

In elementary algebra, the quadratic formula is a formula that provides the solutions to a quadratic equation. Other ways of solving a quadratic equation, such as completing the square, yield the same solutions.

Given a general quadratic equation of the form $ax^{2}+bx+c=0,$ with $x$ representing an unknown, and coefficients $a,$ $b,$ and $c$ representing known real or complex numbers with $a\neq 0,$ the values of $x$ satisfying the equation, called the roots or zeros, can be found using the quadratic formula,

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

where the plus–minus symbol "$\pm$" indicates that the equation has two roots. Written separately, these are:

$$x_{1}={\frac {-b+{\sqrt {b^{2}-4ac}}}{2a}},\qquad x_{2}={\frac {-b-{\sqrt {b^{2}-4ac}}}{2a}}.$$

The quantity $\Delta =b^{2}-4ac$ is known as the discriminant of the quadratic equation. If the coefficients $a,$ $b,$ and $c$ are real numbers then

• when $b^{2}-4ac>0,$ the equation has two distinct real roots;
• when $b^{2}-4ac=0,$ the equation has one repeated real root; and
• when $b^{2}-4ac<0,$ the equation has two distinct complex roots, which are complex conjugates of each other.

Geometrically, the roots represent the $x$ values at which the graph of the function $y=ax^{2}+bx+c,$ a parabola, crosses the $x$-axis. The quadratic formula can also be used to identify the parabola's axis of symmetry.

Equivalent formulations

The quadratic formula can equivalently be written using various alternative expressions, for instance

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

which can be derived by first dividing a quadratic equation by $2a,$ resulting in ${\tfrac {1}{2}}x^{2}+{\tfrac {b}{2a}}x+{\tfrac {c}{2a}}=0,$ then substituting the new coefficients into the standard quadratic formula. Because this variant allows re-use of the intermediately calculated quantity ${\tfrac {b}{2a}},$ it can slightly reduce the arithmetic involved.

Square root in the denominator

A lesser known quadratic formula describes the same roots via an equation with the square root in the denominator (assuming $c\neq 0$):

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

Here the minus–plus symbol "$\mp$" indicates that the two roots of the quadratic equation, in the same order as the standard quadratic formula, are

$$x_{1}={\frac {2c}{-b-{\sqrt {b^{2}-4ac}}}},\qquad x_{2}={\frac {2c}{-b+{\sqrt {b^{2}-4ac}}}}.$$

When $-b$ has the opposite sign as either ${\textstyle +{\sqrt {b^{2}-4ac}}}$ or ${\textstyle -{\sqrt {b^{2}-4ac}},}$ subtraction can cause catastrophic cancellation, resulting in poor accuracy in numerical calculations; choosing between the version of the quadratic formula with the square root in the numerator or denominator depending on the sign of $b$ can avoid this problem. See § Numerical calculation below.

This version of the quadratic formula is used in Muller's method for finding the roots of general functions. It can be derived from the standard formula from the identity $x_{1}x_{2}=c/a,$ one of Vieta's formulas.

Derivations of the formula

Many different methods to derive the quadratic formula are available in the literature. The standard one is a simple application of the completing the square technique. Alternative methods are sometimes simpler than completing the square, and may offer interesting insight into other areas of mathematics.

By completing the square

We start by dividing the quadratic equation ${\textstyle ax^{2}+bx+c=0}$ by the quadratic coefficient $a$, which is allowed because $a$ is non-zero, and then we subtract the constant term $c/a$ to isolate it on the right-hand side:

$${\begin{aligned}ax^{2}+bx+c&=0\\[3mu]x^{2}+{\frac {b}{a}}x+{\frac {c}{a}}&=0\\[3mu]x^{2}+{\frac {b}{a}}x&=-{\frac {c}{a}}.\end{aligned}}$$

The left-hand side is now ready for the method of completing the square, i.e. adding a constant $k^{2}$ to a quadratic polynomial of the form $x^{2}+2kx$ to obtain a squared binomial $x^{2}+2kx+k^{2}={}$$(x+k)^{2}.$ In this example we add ${\textstyle (b/2a)^{2}}$ to both sides so that the left-hand side can be factored:

$${\begin{aligned}x^{2}+2\left({\frac {b}{2a}}\right)x+\left({\frac {b}{2a}}\right)^{2}&=-{\frac {c}{a}}+\left({\frac {b}{2a}}\right)^{2}\\[5mu]\left(x+{\frac {b}{2a}}\right)^{2}&={\frac {b^{2}-4ac}{4a^{2}}}.\end{aligned}}$$

Because the left-hand side is now a perfect square, we can easily take the square root of both sides:

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

Finally, subtracting $b/2a$ from both sides to isolate $x$ produces the quadratic formula:

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

Shorter method

Instead of dividing by $a$ to isolate $x^{2},$ it can be slightly simpler to multiply by $4a$ instead to produce $(2ax)^{2},$ which allows us to complete the square without need for fractions. Then the steps of the derivation are:

1. Multiply each side by $4a.$
2. Add $b^{2}-4ac$ to both sides to complete the square.
3. Take the square root of both sides.
4. Isolate $x.$

Thus, the quadratic formula is derived as follows:

$${\begin{aligned}ax^{2}+bx+c&=0\\[3mu]4a^{2}x^{2}+4abx+4ac&=0\\[3mu]4a^{2}x^{2}+4abx+b^{2}&=b^{2}-4ac\\[3mu](2ax+b)^{2}&=b^{2}-4ac\\[3mu]2ax+b&=\pm {\sqrt {b^{2}-4ac}}\\[5mu]x&={\dfrac {-b\pm {\sqrt {b^{2}-4ac}}}{2a}}.{\vphantom {\bigg )}}\end{aligned}}$$

This derivation of the quadratic formula is ancient and was known to the 8th century Indian mathematician Śrīdhara. Compared with the derivation in standard usage, this alternate derivation avoids fractions and squared fractions until the last step and hence does not require a rearrangement after step 3 to obtain a common denominator in the right side.

By substitution

Another derivation uses a change of variables to put the equation into the form $u^{2}=s$ in terms of a new variable $u$ and some constant expression $s,$ whose roots are then $u=\pm {\sqrt {s}}.$

By substituting $x=u-{\tfrac {b}{2a}}$ into $ax^{2}+bx+c=0,$ expanding the products and combining like terms, and then solving for $u^{2},$ we have:

$${\begin{aligned}a\left(u-{\frac {b}{2a}}\right)^{2}+b\left(u-{\frac {b}{2a}}\right)+c&=0\\[5mu]a\left(u^{2}-{\frac {b}{a}}u+{\frac {b^{2}}{4a^{2}}}\right)+b\left(u-{\frac {b}{2a}}\right)+c&=0\\[5mu]au^{2}-bu+{\frac {b^{2}}{4a}}+bu-{\frac {b^{2}}{2a}}+c&=0\\[5mu]au^{2}+{\frac {4ac-b^{2}}{4a}}&=0\\[5mu]u^{2}&={\frac {b^{2}-4ac}{4a^{2}}}.\end{aligned}}$$

Finally, after taking a square root of both sides and substituting the resulting expression for $u$ back into $x=u-{\tfrac {b}{2a}},$ the familiar quadratic formula emerges:

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

By using algebraic identities

The following method was used by many historical mathematicians:

Let the roots of the quadratic equation $ax^{2}+bx+c=0$ be $\alpha$ and $\beta .$ The derivation starts from an identity for the square of a difference (valid for any two complex numbers), of which we can take the square root on both sides:

$${\begin{aligned}(\alpha -\beta )^{2}&=(\alpha +\beta )^{2}-4\alpha \beta \\[3mu]\alpha -\beta &=\pm {\sqrt {(\alpha +\beta )^{2}-4\alpha \beta }}.\end{aligned}}$$

Since the coefficient $a\neq 0,$ we can divide the quadratic equation by $a$ to obtain a monic polynomial with the same roots. Namely,

$$x^{2}+{\frac {b}{a}}x+{\frac {c}{a}}=(x-\alpha )(x-\beta )=x^{2}-(\alpha +\beta )x+\alpha \beta .$$

This implies that the sum $\alpha +\beta =-{\tfrac {b}{a}}$ and the product $\alpha \beta ={\tfrac {c}{a}}.$ Thus the identity can be rewritten:

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

Therefore,

$${\begin{aligned}\alpha &={\tfrac {1}{2}}(\alpha +\beta )+{\tfrac {1}{2}}(\alpha -\beta )=-{\frac {b}{2a}}\pm {\frac {\sqrt {b^{2}-4ac}}{2a}},\\[10mu]\beta &={\tfrac {1}{2}}(\alpha +\beta )-{\tfrac {1}{2}}(\alpha -\beta )=-{\frac {b}{2a}}\mp {\frac {\sqrt {b^{2}-4ac}}{2a}}.\end{aligned}}$$

The two possibilities for each of $\alpha$ and $\beta$ are the same two roots in opposite order, so we can combine them into the standard quadratic equation:

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

By Lagrange resolvents

Further information: 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. 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 themselves rather than algebraically rearranging the original equation. Given a monic quadratic polynomial $x^{2}+px+q$ assume that $\alpha$ and $\beta$ are the two roots. So the polynomial factors as

$${\begin{aligned}x^{2}+px+q&=(x-\alpha )(x-\beta )\\&=x^{2}-(\alpha +\beta )x+\alpha \beta \end{aligned}}$$

which implies $p=-(\alpha +\beta )$ and $q=\alpha \beta .$

Since multiplication and addition are both commutative, exchanging the roots $\alpha$ and $\beta$ will not change the coefficients $p$ and $q$: one can say that $p$ and $q$ are symmetric polynomials in $\alpha$ and $\beta .$ Specifically, they are the elementary symmetric polynomials – any symmetric polynomial in $\alpha$ and $\beta$ can be expressed in terms of $\alpha +\beta$ and $\alpha \beta$ instead.

The Galois theory approach to analyzing and solving polynomials is to ask whether, given coefficients of a polynomial each of which is a symmetric function in the roots, one can "break" the symmetry and thereby recover the roots. Using this approach, solving a polynomial of degree $n$ is related to the ways of rearranging ("permuting") $n$ terms, called the symmetric group on $n$ letters and denoted $S_{n}.$ For the quadratic polynomial, the only ways to rearrange two roots are to either leave them be or to transpose them, so solving a quadratic polynomial is simple.

To find the roots $\alpha$ and $\beta ,$ consider their sum and difference:

$$r_{1}=\alpha +\beta ,\quad r_{2}=\alpha -\beta .$$

These are called the Lagrange resolvents of the polynomial, from which the roots can be recovered as

$$\alpha ={\tfrac {1}{2}}(r_{1}+r_{2}),\quad \beta ={\tfrac {1}{2}}(r_{1}-r_{2}).$$

Because $r_{1}=\alpha +\beta$ is a symmetric function in $\alpha$ and $\beta ,$ it can be expressed in terms of $p$ and $q,$ specifically $r_{1}=-p$ as described above. However, $r_{2}=\alpha -\beta$ is not symmetric, since exchanging $\alpha$ and $\beta$ yields the additive inverse $-r_{2}=\beta -\alpha .$ So $r_{2}$ cannot be expressed in terms of the symmetric polynomials. However, its square $r_{2}^{2}=(\alpha -\beta )^{2}$ is symmetric in the roots, expressible in terms of $p$ and $q.$ Specifically $r_{2}^{2}=(\alpha -\beta )^{2}={}$$(\alpha +\beta )^{2}-4\alpha \beta ={}$$p^{2}-4q,$ which implies ${\textstyle r_{2}=\pm {\sqrt {p^{2}-4q}}.}$ Taking the positive root "breaks" the symmetry, resulting in

$$r_{1}=-p,\qquad r_{2}={\textstyle {\sqrt {p^{2}-4q}}}$$

from which the roots $\alpha$ and $\beta$ are recovered as

$$x={\tfrac {1}{2}}(r_{1}\pm r_{2})={\tfrac {1}{2}}{\Bigl (}{-p}\pm {\textstyle {\sqrt {p^{2}-4q}}}{\Bigr )}$$

which is the quadratic formula for a monic polynomial.

Substituting $p=b/a,$ $q=c/a$ yields the usual expression for an arbitrary quadratic polynomial. The resolvents can be recognized as

$${\tfrac {1}{2}}r_{1}=-{\tfrac {1}{2}}p=-{\frac {b}{2a}},\qquad r_{2}^{2}=p_{2}-4q={\frac {b^{2}-4ac}{a^{2}}},$$

respectively the vertex and the discriminant of the monic polynomial.

A similar but more complicated method works for cubic equations, which have 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. 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.

Numerical calculation

The quadratic formula is exactly correct when performed using the idealized arithmetic of real numbers, but when approximate arithmetic is used instead, for example pen-and-paper arithmetic carried out to a fixed number of decimal places or the floating-point binary arithmetic available on computers, the limited precision of intermediate calculations can lead to substantially inaccurate results.

When $-b\approx {\sqrt {b^{2}-4ac}},$ evaluation of ${\textstyle -b-{\sqrt {b^{2}-4ac}}}$ causes catastrophic cancellation, as does the evaluation of ${\textstyle -b+{\sqrt {b^{2}-4ac}}}$ when $-b\approx -{\sqrt {b^{2}-4ac}}.$ When using the standard quadratic formula, calculating one of the two roots always involves addition, which preserves the working precision of the intermediate calculations, while calculating the other root involves subtraction, which compromises it. Therefore, naïvely following the standard quadratic formula to solve arbitrary quadratic equations is an unacceptable generic strategy if results should always be accurate. Unfortunately, introductory algebra textbooks typically do not address this problem, even though it causes students to obtain inaccurate results in other school subjects such as introductory chemistry.

For example, if trying to solve the equation ${\textstyle x^{2}-1634x+2=0}$ using a pocket calculator, the result of the quadratic formula ${\textstyle x=817\pm {\sqrt {667\,487}}}$ might be approximately calculated as:

$${\begin{alignedat}{3}x_{1}&=817+816.998\,776\,0&&=1.633\,998\,776\times 10^{3},\\x_{2}&=817-816.998\,776\,0&&=1.224\times 10^{-3}.\end{alignedat}}$$

Even though the calculator used ten decimal digits of precision for each step, calculating the difference between two approximately equal numbers has yielded a result for $x_{2}$ with only four correct digits.

One way to recover an accurate result is to use the identity $x_{1}x_{2}=c/a.$ In this example $x_{2}$ can be calculated as $x_{2}=2/x_{1}={}$$1.223\,991\,125\times 10^{-3},$ which is correct to the full ten digits. Another more or less equivalent approach is to use the version of the quadratic formula with the square root in the denominator to calculate one of the roots (see § Square root in the denominator above).

Practical computer implementations of the solution of quadratic equations commonly choose which formula to use for each root depending on the sign of $b.$

Historical development

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

The Greek mathematician Euclid (circa 300 BC) used geometric methods to solve quadratic equations in Book 2 of his Elements, an influential mathematical treatise. Rules for quadratic equations appear in the Chinese The Nine Chapters on the Mathematical Art circa 200 BC. In his work Arithmetica, the Greek mathematician Diophantus (circa 250 AD) solved quadratic equations with a method more recognizably algebraic than the geometric algebra of Euclid. His solution gives only one root, even when both roots are positive.

The Indian mathematician Brahmagupta (597–668 AD) explicitly described the quadratic formula in his treatise Brāhmasphuṭasiddhānta published in 628 AD, but written in words instead of symbols. 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." In modern notation, this can be written ${\textstyle x={\bigl (}{\sqrt {c\cdot 4a+b^{2}}}-b{\bigr )}{\big /}2a.}$ Śrīdhara (8th century), an Indian mathematician also came up with a similar algorithm for solving quadratic equations, though there is no indication that he considered both the roots.

The 9th-century Persian mathematician Muḥammad ibn Mūsā al-Khwārizmī solved quadratic equations algebraically. The quadratic formula covering all cases was first obtained by Simon Stevin in 1594. In 1637 René Descartes published La Géométrie containing special cases of the quadratic formula in the form we know today.

Geometric 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

In terms of coordinate geometry, an axis-aligned parabola is a curve whose $(x,y)$-coordinates are the graph of a second-degree polynomial, of the form $y=ax^{2}+bx+c,$ where $a,$ $b,$ and $c$ are real-valued constant coefficients with $a\neq 0.$

Geometrically, the quadratic formula defines the points $(x,0)$ on the graph, where the parabola crosses the $x$-axis. Furthermore, it can be separated into two terms,

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

The first term describes the axis of symmetry, the line $x=-{\tfrac {b}{2a}}.$ The second term, ${\sqrt {b^{2}-4ac}}{\big /}2a,$ gives the distance the roots are away from the axis of symmetry.

If the parabola's vertex is on the $x$-axis, then the corresponding equation has a single repeated root on the line of symmetry, and this distance term is zero; algebraically, the discriminant $b^{2}-4ac=0.$

If the discriminant is positive, then the vertex is not on the $x$-axis but the parabola opens in the direction of the $x$-axis, crossing it twice, so the corresponding equation has two real roots. If the discriminant is negative, then the parabola opens in the opposite direction, never crossing the $x$-axis, and the equation has no real roots; in this case the two complex-valued roots will be complex conjugates whose real part is the $x$ value of the axis of symmetry.

Dimensional analysis

If the constants $a,$ $b,$ and/or $c$ are not unitless then the quantities $x$ and ${\tfrac {b}{a}}$ must have the same units, because the terms $ax^{2}$ and $bx$ agree on their units. By the same logic, the coefficient $c$ must have the same units as ${\tfrac {b^{2}}{a}},$ irrespective of the units of $x.$ This can be a powerful tool for verifying that a quadratic expression of physical quantities has been set up correctly.

See also

• Fundamental theorem of algebra
• Vieta's formulas