Completing the square

From Wikipedia, the free encyclopedia

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

$ax^2 + bx + c\,\!$

to the form

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

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

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

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
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 External links

[edit] Overview

[edit] Background

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

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

For example:

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

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

[edit] Basic example

Consider the following quadratic polynomial:

$x^2 + 10x + 28.\,\!$

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

$(x+5)^2 \,=\, x^2 + 10x + 25.\,\!$

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

$x^2 + 10x + 28 \,=\, (x+5)^2 + 3.$

This is called completing the square.

[edit] General description

Given any quadratic of the form

$x^2 + bx + c,\,\!$

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

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

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

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

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

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

[edit] Non-monic case

Given a quadratic polynomial of the form

$ax^2 + bx + c\,\!$

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

$3x^2 + 12x + 27 \,=\, 3(x^2+4x+9) \,=\, 3\left((x+2)^2 + 5\right) \,=\, 3(x+2)^2 + 15.$

This allows us to write any quadratic polynomial in the form

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

[edit] Formula

The result of completing the square may be written as a formula. Specifically,

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

For the general case:

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

[edit] 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

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

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.

[edit] Solving quadratic equations

Completing the square may be used to solve any quadratic equation. Given a quadratic equation

$x^2 + 6x + 5 = 0,\,\!$

the first step is to complete the square:

$(x+3)^2 - 4 = 0.\,\!$

Next we solve for the squared term:

$(x+3)^2 = 4.\,\!$

Then either

$x+3 = -2 \quad\text{or}\quad x+3 = 2,$

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.) Indeed, this method can be used to derive the quadratic formula for the roots of a general quadratic polynomial (see Quadratic equation#Derivation).

[edit] Irrational roots

Unlike methods involving factoring the constant term, completing the square will find the roots of a quadratic equation even when those roots are irrational. For example, consider the equation

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

Completing the square gives

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

so

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

Then either

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

and therefore

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

[edit] Complex roots

Completing the square can also find roots of a quadratic equation that are complex numbers:

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

[edit] Non-monic case

For an equation involving a non-monic quadratic, simply start by dividing through by the coefficient of x². For example:

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

[edit] Other applications

[edit] Integration

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

$\int\frac{dx}{ax^2+bx+c}$

using the basic integrals

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

For example, consider the integral

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

Completing the square in the denominator gives:

$\int\frac{dx}{(x+3)^2 + 4} \,=\, \int\frac{dx}{(x+3)^2 + 2^2}.$

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

$\int\frac{dx}{(x+3)^2 + 4} \,=\, \frac{1}{2}\arctan\left(\frac{x+3}{2}\right)$

[edit] Complex numbers

Consider the expression

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

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

$|z-b|^2 - |b|^2 + c , \,\!$

which is clearly a real quantity. This is because

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

As another example, the expression

$ax^2 + by^2 + c , \,\!$

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

$z = \sqrt{a}\,x + i \sqrt{b} \,y .$

Then

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

so

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

[edit] Geometric perspective

Consider completing the square for the equation

$x^2 + bx = a.\,$

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]

[edit] A variation on the technique

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

$u^2 + 2uv\,$

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

$u^2 + v^2\,$

to get a square.

[edit] Example: the sum of a positive number and its reciprocal

By writing

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

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.

[edit] Example: factoring a simple quartic polynomial

Consider the problem of factoring the polynomial

$x^4 + 324 . \,\!$

This is

$(x^2)^2 + (18)^2, \,\!$

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

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

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

[edit] External links

Completing the square on PlanetMath

Completing the square

From Wikipedia, the free encyclopedia

Contents

[edit] Overview

[edit] Background

[edit] Basic example

[edit] General description

[edit] Non-monic case

[edit] Formula

[edit] Relation to the graph

[edit] Solving quadratic equations

[edit] Irrational roots

[edit] Complex roots

[edit] Non-monic case

[edit] Other applications

[edit] Integration

[edit] Complex numbers

[edit] Geometric perspective

[edit] A variation on the technique

[edit] Example: the sum of a positive number and its reciprocal

[edit] Example: factoring a simple quartic polynomial

[edit] External links

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