Completing the square

In elementary algebra, completing the square is a technique for converting a quadratic polynomial of the form

$ax^2 %2B bx %2B c\,\!$

to the form

$a(\cdots\cdots)^2 %2B \mbox{constant}.\,$

In this context, "constant" means not depending on x. The expression inside the parenthesis is of the form (x − constant). Thus one converts ax² + bx + c to

$a(x - h)^2 %2B k\,$

and one must find h and k.

Completing the square is used in

solving quadratic equations,
graphing quadratic functions,
evaluating integrals in calculus,
finding Laplace transforms.

In mathematics, completing the square is considered a basic algebraic operation, and is often applied without remark in any computation involving quadratic polynomials.

1 Overview
2 Relation to the graph
3 Solving quadratic equations
- 3.1 Irrational and complex roots
- 3.2 Non-monic case
4 Other applications
- 4.1 Integration
- 4.2 Complex numbers
5 Geometric perspective
6 A variation on the technique
- 6.1 Example: the sum of a positive number and its reciprocal
- 6.2 Example: factoring a simple quartic polynomial
7 References
8 External links

Overview

Background

There is a simple formula in elementary algebra for computing the square of a binomial:

$(x %2B p)^2 \,=\, x^2 %2B 2px %2B p^2.\,\!$

For example:

$\begin{alignat}{2} (x%2B3)^2 \,&=\, x^2 %2B 6x %2B 9 && (p=3)\\[3pt] (x-5)^2 \,&=\, x^2 - 10x %2B 25\qquad && (p=-5). \end{alignat}$

In any perfect square, the number p is always half the coefficient of x, and the constant term is equal to p².

Basic example

Consider the following quadratic polynomial:

$x^2 %2B 10x %2B 28.\,\!$

This quadratic is not a perfect square, since 28 is not the square of 5:

$(x%2B5)^2 \,=\, x^2 %2B 10x %2B 25.\,\!$

However, it is possible to write the original quadratic as the sum of this square and a constant:

$x^2 %2B 10x %2B 28 \,=\, (x%2B5)^2 %2B 3.$

This is called completing the square.

General description

Given any monic quadratic

$x^2 %2B bx %2B c,\,\!$

it is possible to form a square that has the same first two terms:

$\left(x%2B\tfrac{1}{2} b\right)^2 \,=\, x^2 %2B bx %2B \tfrac{1}{4}b^2.$

This square differs from the original quadratic only in the value of the constant term. Therefore, we can write

$x^2 %2B bx %2B c \,=\, \left(x %2B \tfrac{1}{2}b\right)^2 %2B k,$

where k is a constant. This operation is known as completing the square. For example:

$\begin{alignat}{1} x^2 %2B 6x %2B 11 \,&=\, (x%2B3)^2 %2B 2 \\[3pt] x^2 %2B 14x %2B 30 \,&=\, (x%2B7)^2 - 19 \\[3pt] x^2 - 2x %2B 7 \,&=\, (x-1)^2 %2B 6. \end{alignat}$

Non-monic case

Given a quadratic polynomial of the form

$ax^2 %2B bx %2B c\,\!$

it is possible to factor out the coefficient a, and then complete the square for the resulting monic polynomial.

Example:

$\begin{align} 3x^2 %2B 12x %2B 27 &= 3(x^2%2B4x%2B9)\\ &{}= 3\left((x%2B2)^2 %2B 5\right)\\ &{}= 3(x%2B2)^2 %2B 15 \end{align}$

This allows us to write any quadratic polynomial in the form

$a(x-h)^2 %2B k.\,\!$

Formula

The result of completing the square may be written as a formula. For the general case:^[1]

$ax^2 %2B bx %2B c \;=\; a(x-h)^2 %2B k,\quad\text{where}\quad h = -\frac{b}{2a} \quad\text{and}\quad k = c - \frac{b^2}{4a}.$

Specifically, when a=1:

$x^2 %2B bx %2B c \;=\; (x-h)^2 %2B k,\quad\text{where}\quad h = -\frac{b}{2} \quad\text{and}\quad k = c - \frac{b^2}{4}.$

Relation to the graph

In analytic geometry, the graph of any quadratic function is a parabola in the xy-plane. Given a quadratic polynomial of the form

$(x-h)^2 %2B k \quad\text{or}\quad a(x-h)^2 %2B k$

the numbers h and k may be interpreted as the Cartesian coordinates of the vertex of the parabola. That is, h is the x-coordinate of the axis of symmetry, and k is the minimum value (or maximum value, if a < 0) of the quadratic function.

In other words, the graph of the function ƒ(x) = x² is a parabola whose vertex is at the origin (0, 0). Therefore, the graph of the function ƒ(x − h) = (x − h)² is a parabola shifted to the right by h whose vertex is at (h, 0), as shown in the top figure. In contrast, the graph of the function ƒ(x) + k = x² + k is a parabola shifted upward by k whose vertex is at (0, k), as shown in the center figure. Combining both horizontal and vertical shifts yields ƒ(x − h) + k = (x − h)² + k is a parabola shifted to the right by h and upward by k whose vertex is at (h, k), as shown in the bottom figure.

Solving quadratic equations

Completing the square may be used to solve any quadratic equation. For example:

$x^2 %2B 6x %2B 5 = 0,\,\!$

The first step is to complete the square:

$(x%2B3)^2 - 4 = 0.\,\!$

Next we solve for the squared term:

$(x%2B3)^2 = 4.\,\!$

Then either

$x%2B3 = -2 \quad\text{or}\quad x%2B3 = 2,$

and therefore

$x = -5 \quad\text{or}\quad x = -1.$

This can be applied to any quadratic equation. When the x² has a coefficient other than 1, the first step is to divide out the equation by this coefficient: for an example see the non-monic case below.

Irrational and complex roots

Unlike methods involving factoring the equation, which is only reliable if the roots are rational, completing the square will find the roots of a quadratic equation even when those roots are irrational or complex. For example, consider the equation

$x^2 - 10x %2B 18 = 0.\,\!$

Completing the square gives

$(x-5)^2 - 7 = 0,\,\!$

$(x-5)^2 = 7.\,\!$

Then either

$x-5 = -\sqrt{7} \quad\text{or}\quad x-5 = \sqrt{7},\,$

$x = 5 - \sqrt{7}\quad\text{or}\quad x = 5 %2B \sqrt{7}. \,$

In terser language:

$x = 5 \pm \sqrt{7}.\,$

Equations with complex roots can be handled in the same way. For example:

$\begin{array}{c} x^2 %2B 4x %2B 5 \,=\, 0 \\[6pt] (x%2B2)^2 %2B 1 \,=\, 0 \\[6pt] (x%2B2)^2 \,=\, -1 \\[6pt] x%2B2 \,=\, \pm i \\[6pt] x \,=\, -2 \pm i. \end{array}$

Non-monic case

For an equation involving a non-monic quadratic, the first step to solving them is to divide through by the coefficient of x². For example:

$\begin{array}{c} 2x^2 %2B 7x %2B 6 \,=\, 0 \\[6pt] x^2 %2B \tfrac{7}{2}x %2B 3 \,=\, 0 \\[6pt] \left(x%2B\tfrac{7}{4}\right)^2 - \tfrac{1}{16} \,=\, 0 \\[6pt] \left(x%2B\tfrac{7}{4}\right)^2 \,=\, \tfrac{1}{16} \\[6pt] x%2B\tfrac{7}{4} = \tfrac{1}{4} \quad\text{or}\quad x%2B\tfrac{7}{4} = -\tfrac{1}{4} \\[6pt] x = -\tfrac{3}{2} \quad\text{or}\quad x = -2. \end{array}$

Other applications

Integration

Completing the square may be used to evaluate any integral of the form

$\int\frac{dx}{ax^2%2Bbx%2Bc}$

using the basic integrals

$\int\frac{dx}{x^2 - a^2} = \frac{1}{2a}\ln\left|\frac{x-a}{x%2Ba}\right| %2BC \quad\text{and}\quad \int\frac{dx}{x^2 %2B a^2} = \frac{1}{a}\arctan\left(\frac{x}{a}\right) %2BC.$

For example, consider the integral

$\int\frac{dx}{x^2 %2B 6x %2B 13}.$

Completing the square in the denominator gives:

$\int\frac{dx}{(x%2B3)^2 %2B 4} \,=\, \int\frac{dx}{(x%2B3)^2 %2B 2^2}.$

This can now be evaluated by using the substitution u = x + 3, which yields

$\int\frac{dx}{(x%2B3)^2 %2B 4} \,=\, \frac{1}{2}\arctan\left(\frac{x%2B3}{2}\right)%2BC.$

Complex numbers

Consider the expression

$|z|^2 - b^*z - bz^* %2B c,\,$

where z and b are complex numbers, z^* and b^* are the complex conjugates of z and b, respectively, and c is a real number. Using the identity |u|² = uu^* we can rewrite this as

$|z-b|^2 - |b|^2 %2B c , \,\!$

which is clearly a real quantity. This is because

$\begin{align} |z-b|^2 &{}= (z-b)(z-b)^*\\ &{}= (z-b)(z^*-b^*)\\ &{}= zz^* - zb^* - bz^* %2B bb^*\\ &{}= |z|^2 - zb^* - bz^* %2B |b|^2 . \end{align}$

As another example, the expression

$ax^2 %2B by^2 %2B c , \,\!$

where a, b, c, x, and y are real numbers, with a > 0 and b > 0, may be expressed in terms of the square of the absolute value of a complex number. Define

$z = \sqrt{a}\,x %2B i \sqrt{b} \,y .$

Then

$\begin{align} |z|^2 &{}= z z^*\\ &{}= (\sqrt{a}\,x %2B i \sqrt{b}\,y)(\sqrt{a}\,x - i \sqrt{b}\,y) \\ &{}= ax^2 - i\sqrt{ab}\,xy %2B i\sqrt{ba}\,yx - i^2by^2 \\ &{}= ax^2 %2B by^2 , \end{align}$

$ax^2 %2B by^2 %2B c = |z|^2 %2B c . \,\!$

Geometric perspective

Consider completing the square for the equation

$x^2 %2B bx = a.\,$

Since x² represents the area of a square with side of length x, and bx represents the area of a rectangle with sides b and x, the process of completing the square can be viewed as visual manipulation of rectangles.

Simple attempts to combine the x² and the bx rectangles into a larger square result in a missing corner. The term (b/2)² added to each side of the above equation is precisely the area of the missing corner, whence derives the terminology "completing the square". [1]

A variation on the technique

As conventionally taught, completing the square consists of adding the third term, v² to

$u^2 %2B 2uv\,$

to get a square. There are also cases in which one can add the middle term, either 2uv or −2uv, to

$u^2 %2B v^2\,$

to get a square.

Example: the sum of a positive number and its reciprocal

By writing

$\begin{align} x %2B {1 \over x} &{} = \left(x - 2 %2B {1 \over x}\right) %2B 2\\ &{}= \left(\sqrt{x} - {1 \over \sqrt{x}}\right)^2 %2B 2 \end{align}$

we show that the sum of a positive number x and its reciprocal is always greater than or equal to 2. The square of a real expression is always greater than or equal to zero, which gives the stated bound; and here we achieve 2 just when x is 1, causing the square to vanish.

Example: factoring a simple quartic polynomial

Consider the problem of factoring the polynomial

$x^4 %2B 324 . \,\!$

This is

$(x^2)^2 %2B (18)^2, \,\!$

so the middle term is 2(x²)(18) = 36x². Thus we get

$\begin{align} x^4 %2B 324 &{}= (x^4 %2B 36x^2 %2B 324 ) - 36x^2 \\ &{}= (x^2 %2B 18)^2 - (6x)^2 =\text{a difference of two squares} \\ &{}= (x^2 %2B 18 %2B 6x)(x^2 %2B 18 - 6x) \\ &{}= (x^2 %2B 6x %2B 18)(x^2 - 6x %2B 18) \end{align}$

(the last line being added merely to follow the convention of decreasing degrees of terms).

References

^ Narasimhan, Revathi (2008). Precalculus: Building Concepts and Connections. Cengage Learning. p. 133–134. ISBN 0-618-41301-4. http://books.google.com/books?id=hLZz3xcP0SAC. , Section Formula for the Vertex of a Quadratic Function, page 133–134, figure 2.4.8

Algebra 1, Glencoe, ISBN 0-07-825083-8, pages 539–544
Algebra 2, Saxon, ISBN 0-939798-62-X, pages 214–214, 241–242, 256–257, 398–401

External links

Completing the square on PlanetMath

Completing the square

Contents

Overview

Background

Basic example

General description

Non-monic case

Formula

Relation to the graph

Solving quadratic equations

Irrational and complex roots

Non-monic case

Other applications

Integration

Complex numbers

Geometric perspective

A variation on the technique

Example: the sum of a positive number and its reciprocal

Example: factoring a simple quartic polynomial

References

External links