Completing the square


Completing the square

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

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

Contents

Overview

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 the constant term is equal to p2.

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.


General description

Given any monic quadratic

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}

Non-monic case

Given a quadratic polynomial of the form

ax^2 + bx + 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 + 12x + 27 &= 3(x^2+4x+9)\\
          &{}= 3\left((x+2)^2 + 5\right)\\
          &{}= 3(x+2)^2 + 15
\end{align}

This allows us to write any quadratic polynomial in the form

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

Formula

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

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}.

Specifically, when a=1:

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}.

Relation to the graph

Graphs of quadratic functions shifted to the right by h = 0, 5, 10, and 15.
Graphs of quadratic functions shifted upward by k = 0, 5, 10, and 15.
Graphs of quadratic functions shifted upward and to the right by 0, 5, 10, and 15.

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.

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) + kx2 + 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 (hk), as shown in the bottom figure.

Solving quadratic equations

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

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: 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 + 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},\,

so

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

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 + 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}

Other applications

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

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)+C.

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 . \,\!

Geometric perspective

Completing the square 307.PNG

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]

A variation on the technique

As conventionally taught, completing the square consists of adding the third term, v 2 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.

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.

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).

References

External links


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