finding functionality2: Completing the Square by Professor Davis

Today I'm going to teach everyone how to complete the square and what the concept really means.

n 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

$ax^2 + bx + c\,\!$ is converted to

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

for some values of h and k.
Completing the square is used in

solving quadratic equations,
graphing quadratic functions,
evaluating integrals in calculus, such as Gaussian integrals with a linear term in the exponent
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. Completing the square is also used to derive the quadratic formula.

There are many different formulas for completing the square and here they are....

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

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

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) + 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.

I hope you all understand completing the square a little better after reading my lesson. If not take a look at this very informational YouTube clip. This is just another way of explaining what i have already discussed.

https://www.youtube.com/watch?v=gzm-uhj06q8

4 comments:

AnonymousApril 11, 2014 at 4:25 PM
You went really in depth and all of the examples provided a lot of detail which explained all of your points. I really liked all of it especially because I think the quadratic function is one of my favorite things to do!
AnonymousApril 11, 2014 at 4:41 PM
Very in depth response that really helps understand the idea of completing the square!
AnonymousApril 22, 2014 at 11:47 AM
Thanks for this, it really helped me as I have had difficulties in the past with completing the square. I feel much more confident about it now. Your pictures and descriptions helped a lot.
lilnesscapadesMay 4, 2014 at 6:09 PM
jonathan,

this is a very theoretically heavy lesson that you took on. some simpler examples may have been better, especially you showing step by step "how" to complete the square. the transformation examples were very nice.

professor little

finding functionality2

image

Friday, April 11, 2014

Completing the Square by Professor Davis