Quadratic equation

It may be possible to express a quadratic equation ax² + bx + c = 0 as a product (px + q)(rx + s) = 0. In some cases, it is possible, by simple inspection, to determine values of p, q, r, and s that make the two forms equivalent to one another. If the quadratic equation is written in the second form, then the "Zero Factor Property" states that the quadratic equation is satisfied if px + q = 0 or rx + s = 0. Solving these two linear equations provides the roots of the quadratic.

For most students, factoring by inspection is the first method of solving quadratic equations to which they are exposed.[2]^{: 202–207} If one is given a quadratic equation in the form x² + bx + c = 0, the sought factorization has the form (x + q)(x + s), and one has to find two numbers q and s that add up to b and whose product is c (this is sometimes called "Vieta's rule"[3] and is related to Vieta's formulas). As an example, x² + 5x + 6 factors as (x + 3)(x + 2). The more general case where a does not equal 1 can require a considerable effort in trial and error guess-and-check, assuming that it can be factored at all by inspection.

Except for special cases such as where b = 0 or c = 0, factoring by inspection only works for quadratic equations that have rational roots. This means that the great majority of quadratic equations that arise in practical applications cannot be solved by factoring by inspection.[2]^: 207

Figure 2. For the quadratic function y = x² − x − 2, the points where the graph crosses the x-axis, x = −1 and x = 2, are the solutions of the quadratic equation x² − x − 2 = 0.

The process of completing the square makes use of the algebraic identity

${\displaystyle x^{2}+2hx+h^{2}=(x+h)^{2},}$

which represents a well-defined algorithm that can be used to solve any quadratic equation.[2]^: 207 Starting with a quadratic equation in standard form, ax² + bx + c = 0

1. Divide each side by a, the coefficient of the squared term.
2. Subtract the constant term c/a from both sides.
3. Add the square of one-half of b/a, the coefficient of x, to both sides. This "completes the square", converting the left side into a perfect square.
4. Write the left side as a square and simplify the right side if necessary.
5. Produce two linear equations by equating the square root of the left side with the positive and negative square roots of the right side.
6. Solve each of the two linear equations.

We illustrate use of this algorithm by solving 2x² + 4x − 4 = 0

${\displaystyle 2x^{2}+4x-4=0}$

${\displaystyle \ x^{2}+2x-2=0}$

${\displaystyle \ x^{2}+2x=2}$

${\displaystyle \ x^{2}+2x+1=2+1}$

${\displaystyle \left(x+1\right)^{2}=3}$

${\displaystyle \ x+1=\pm {\sqrt {3}}}$

${\displaystyle \ x=-1\pm {\sqrt {3}}}$

The plus–minus symbol "±" indicates that both x = −1 + √3 and x = −1 − √3 are solutions of the quadratic equation.[4]

Completing the square can be used to derive a general formula for solving quadratic equations, called the quadratic formula.[5] The mathematical proof will now be briefly summarized.[6] It can easily be seen, by polynomial expansion, that the following equation is equivalent to the quadratic equation:

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

Taking the square root of both sides, and isolating x, gives:

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

Some sources, particularly older ones, use alternative parameterizations of the quadratic equation such as ax² + 2bx + c = 0 or ax² − 2bx + c = 0 ,[7] where b has a magnitude one half of the more common one, possibly with opposite sign. These result in slightly different forms for the solution, but are otherwise equivalent.

A number of alternative derivations can be found in the literature. These proofs are simpler than the standard completing the square method, represent interesting applications of other frequently used techniques in algebra, or offer insight into other areas of mathematics.

A lesser known quadratic formula, as used in Muller's method provides the same roots via the equation

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

This can be deduced from the standard quadratic formula by Vieta's formulas, which assert that the product of the roots is c/a.

One property of this form is that it yields one valid root when a = 0, while the other root contains division by zero, because when a = 0, the quadratic equation becomes a linear equation, which has one root. By contrast, in this case, the more common formula has a division by zero for one root and an indeterminate form 0/0 for the other root. On the other hand, when c = 0, the more common formula yields two correct roots whereas this form yields the zero root and an indeterminate form 0/0.

It is sometimes convenient to reduce a quadratic equation so that its leading coefficient is one. This is done by dividing both sides by a, which is always possible since a is non-zero. This produces the reduced quadratic equation:[8]

${\displaystyle x^{2}+px+q=0,}$

where p = b/a and q = c/a. This monic polynomial equation has the same solutions as the original.

The quadratic formula for the solutions of the reduced quadratic equation, written in terms of its coefficients, is:

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

or equivalently:

${\displaystyle x=-{\frac {p}{2}}\pm {\sqrt {\left({\frac {p}{2}}\right)^{2}-q}}.}$

Figure 3. Discriminant signs

In the quadratic formula, the expression underneath the square root sign is called the discriminant of the quadratic equation, and is often represented using an upper case D or an upper case Greek delta:[9]

${\displaystyle \Delta =b^{2}-4ac.}$

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

• If the discriminant is positive, then there are two distinct roots

${\displaystyle {\frac {-b+{\sqrt {\Delta }}}{2a}}\quad {\text{and}}\quad {\frac {-b-{\sqrt {\Delta }}}{2a}},}$

both of which are real numbers. For quadratic equations with rational coefficients, if the discriminant is a square number, then the roots are rational—in other cases they may be quadratic irrationals.

• If the discriminant is zero, then there is exactly one real root ${\displaystyle -{\frac {b}{2a}},}$ sometimes called a repeated or double root.

• If the discriminant is negative, then there are no real roots. Rather, there are two distinct (non-real) complex roots[10]${\displaystyle -{\frac {b}{2a}}+i{\frac {\sqrt {-\Delta }}{2a}}\quad {\text{and}}\quad -{\frac {b}{2a}}-i{\frac {\sqrt {-\Delta }}{2a}},}$

which are complex conjugates of each other. In these expressions 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.

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).[11]

The function f(x) = ax² + bx + c is a quadratic function.[12] The graph of any quadratic function has the same general shape, which is called a parabola. The location and size of the parabola, and how it opens, depend on the values of a, b, and c. As shown in Figure 1, if a > 0, the parabola has a minimum point and opens upward. If a < 0, the parabola has a maximum point and opens downward. The extreme point of the parabola, whether minimum or maximum, corresponds to its vertex. The x-coordinate of the vertex will be located at ${\displaystyle \scriptstyle x={\tfrac {-b}{2a}}}$, and the y-coordinate of the vertex may be found by substituting this x-value into the function. The y-intercept is located at the point (0, c).

The solutions of the quadratic equation ax² + bx + c = 0 correspond to the roots of the function f(x) = ax² + bx + c, since they are the values of x for which f(x) = 0. As shown in Figure 2, if a, b, and c are real numbers and the domain of f is the set of real numbers, then the roots of f are exactly the x-coordinates of the points where the graph touches the x-axis. As shown in Figure 3, 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.

The term

${\displaystyle x-r}$

is a factor of the polynomial

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

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

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

It follows from the quadratic formula that

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

In the special case b² = 4ac where the quadratic has only one distinct root (i.e. the discriminant is zero), the quadratic polynomial can be factored as

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

Figure 4. Graphing calculator computation of one of the two roots of the quadratic equation 2x² + 4x − 4 = 0. Although the display shows only five significant figures of accuracy, the retrieved value of xc is 0.732050807569, accurate to twelve significant figures.

A quadratic function without real root: y = (x − 5)² + 9. The "3" is the imaginary part of the x-intercept. The real part is the x-coordinate of the vertex. Thus the roots are 5 ± 3i.

The solutions of the quadratic equation

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

may be deduced from the graph of the quadratic function

${\displaystyle f(x)=ax^{2}+bx+c,}$

which is a parabola.

If the parabola intersects the x-axis in two points, there are two real roots, which are the x-coordinates of these two points (also called x-intercept).

If the parabola is tangent to the x-axis, there is a double root, which is the x-coordinate of the contact point between the graph and parabola.

If the parabola does not intersect the x-axis, there are two complex conjugate roots. Although these roots cannot be visualized on the graph, their real and imaginary parts can be.[13]

Let h and k be respectively the x-coordinate and the y-coordinate of the vertex of the parabola (that is the point with maximal or minimal y-coordinate. The quadratic function may be rewritten

${\displaystyle y=a(x-h)^{2}+k.}$

Let d be the distance between the point of y-coordinate 2k on the axis of the parabola, and a point on the parabola with the same y-coordinate (see the figure; there are two such points, which give the same distance, because of the symmetry of the parabola). Then the real part of the roots is h, and their imaginary part are ±d. That is, the roots are

${\displaystyle h+id\quad {\text{and}}\quad h-id,}$

or in the case of the example of the figure

${\displaystyle 5+3i\quad {\text{and}}\quad 5-3i.}$

Although the quadratic formula provides an exact solution, the result is not exact if real numbers are approximated during the computation, as usual in numerical analysis, where real numbers are approximated by floating point numbers (called "reals" in many programming languages). In this context, the quadratic formula is not completely stable.

This occurs when the roots have different order of magnitude, or, equivalently, when b² and b² − 4ac are close in magnitude. In this case, the subtraction of two nearly equal numbers will cause loss of significance or catastrophic cancellation in the smaller root. To avoid this, the root that is smaller in magnitude, r, can be computed as ${\displaystyle (c/a)/R}$ where R is the root that is bigger in magnitude.

A second form of cancellation can occur between the terms b² and 4ac of the discriminant, that is when the two roots are very close. This can lead to loss of up to half of correct significant figures in the roots.[7][14]

Graph of the difference between Vieta's approximation for the smallest root of the quadratic equation x² + bx + c = 0 compared with the value calculated using the quadratic formula

Vieta's formulas (named after François Viète) are the relations

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

between the roots of a quadratic polynomial and its coefficients. They result from comparing term by the relation

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

with the equation

${\displaystyle x^{2}+{\frac {b}{a}}x+{\frac {c}{a}}=0.}$

The first Vieta's formula is useful for graphing a quadratic function. Since the graph is symmetric with respect to a vertical line through the vertex, the vertex's x-coordinate is located at the average of the roots (or intercepts). Thus the x-coordinate of the vertex is

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

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

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

These formulas for the vertex can also deduced directly from the formula (see Completing the square)

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

For numerical computation, Vieta's formulas provide a useful method for finding the roots of a quadratic equation in the case where one root is much smaller than the other. If |x₂| << |x₁|, then x₁ + x₂ ≈ x₁, and we have the estimate:

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

The second Vieta's formula then provides:

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

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.[30] 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.[31] Calculating complex roots would require using a different trigonometric form.[32]

To illustrate, let us assume we had available seven-place logarithm and trigonometric tables, and wished to solve the following to six-significant-figure accuracy:

${\displaystyle 4.16130x^{2}+9.15933x-11.4207=0}$

1. A seven-place lookup table might have only 100,000 entries, and computing intermediate results to seven places would generally require interpolation between adjacent entries.
2. ${\displaystyle \log a=0.6192290,\log b=0.9618637,\log c=1.0576927}$
3. ${\displaystyle 2{\sqrt {ac}}/b=2\times 10^{(0.6192290+1.0576927)/2-0.9618637}=1.505314}$
4. ${\displaystyle \theta =(\tan ^{-1}1.505314)/2=28.20169^{\circ }{\text{ or }}-61.79831^{\circ }}$
5. ${\displaystyle \log |\tan \theta |=-0.2706462{\text{ or }}0.2706462}$
6. ${\displaystyle \log {\sqrt {c/a}}=(1.0576927-0.6192290)/2=0.2192318}$
7. ${\displaystyle x_{1}=10^{0.2192318-0.2706462}=0.888353}$ (rounded to six significant figures)

${\displaystyle x_{2}=-10^{0.2192318+0.2706462}=-3.08943}$

If the quadratic equation ${\displaystyle ax^{2}+bx+c=0}$ with real coefficients has two complex roots—the case where ${\displaystyle b^{2}-4ac<0,}$ requiring a and c to have the same sign as each other—then the solutions for the roots can be expressed in polar form as[33]

${\displaystyle x_{1},\,x_{2}=r(\cos \theta \pm i\sin \theta ),}$

where ${\displaystyle r={\sqrt {\tfrac {c}{a}}}}$ and ${\displaystyle \theta =\cos ^{-1}\left({\tfrac {-b}{2{\sqrt {ac}}}}\right).}$

Figure 6. Geometric solution of ax² + bx + c = 0 using Lill's method. Solutions are −AX1/SA, −AX2/SA

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 Figure 6. 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.[34]

Carlyle circle of the quadratic equation x² − sx + p = 0.

The Carlyle circle, named after Thomas Carlyle, has the property that the solutions of the quadratic equation are the horizontal coordinates of the intersections of the circle with the horizontal axis.[35] Carlyle circles have been used to develop ruler-and-compass constructions of regular polygons.

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. 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

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

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

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

and there is only one root since

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

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.