# Difference between revisions of "Completing the Square"

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*Step 2: Factor out the coefficient of the squared term <br> | *Step 2: Factor out the coefficient of the squared term <br> | ||

{{Hide|1=<math>a(x^2+\frac{b}{a})=-c</math>}} | {{Hide|1=<math>a(x^2+\frac{b}{a})=-c</math>}} | ||

− | *Step 3:Complete the quadratic in the | + | *Step 3: Complete the quadratic in the parenthesis to make a perfect square trinomial. Remember that in general the constant of the perfect square trinomial is the square of half of the coefficient of x.<br> |

{{Hide|1=<math>a(x^2+\frac{b}{a}+(\frac{b}{2a})^2)=a(x^2+\frac{b}{a}+\frac{b^2}{4a^2})</math>}} | {{Hide|1=<math>a(x^2+\frac{b}{a}+(\frac{b}{2a})^2)=a(x^2+\frac{b}{a}+\frac{b^2}{4a^2})</math>}} | ||

*Step 4: Add the equivalent added to the left side of the equation, to the right side of the equation to maintain the equality. ''Remember to multiply the obtained constant by a when added to the right side''. | *Step 4: Add the equivalent added to the left side of the equation, to the right side of the equation to maintain the equality. ''Remember to multiply the obtained constant by a when added to the right side''. | ||

Line 39: | Line 39: | ||

*Step 5: Factor the left side, simplify the right side and bring it over to the left.<br> | *Step 5: Factor the left side, simplify the right side and bring it over to the left.<br> | ||

{{Hide|1=<math>a(x+\frac{b}{2a})^2+c-\frac{b^2}{4a}=0</math><br><br>}} | {{Hide|1=<math>a(x+\frac{b}{2a})^2+c-\frac{b^2}{4a}=0</math><br><br>}} | ||

+ | *Step 6: If asked for, solve for x. | ||

+ | <template>AlignEquals | ||

+ | |e1l=a(x+\frac{b}{2a})^2+c-\frac{b^2}{4a} | ||

+ | |e1r=0 | ||

+ | |e2l=a(x+\frac{b}{2a})^2 | ||

+ | |e2r=-c+a(\frac{b^2}{4a^2}) | ||

+ | |e3l=(x+\frac{b}{2a})^2 | ||

+ | |e3r=\frac{-c+a(\frac{b^2}{4a^2})}{a} | ||

+ | |e4l=(x+\frac{b}{2a})^2 | ||

+ | |e4r=\frac{-c}{a}+\frac{b^2}{4a^2} | ||

+ | |e5l=(x+\frac{b}{2a} | ||

+ | |e5r | ||

+ | </template> | ||

Though the process of completing the square is finalized, it does not look like the <balloon title="load:vertexform">vertex form equation</balloon> <span id="vertexform" style="display:none"><math>a(x-h)^2+k=0</math></span> | Though the process of completing the square is finalized, it does not look like the <balloon title="load:vertexform">vertex form equation</balloon> <span id="vertexform" style="display:none"><math>a(x-h)^2+k=0</math></span> | ||

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All the different letters are bound to get confusing. It may be easier to visualize this process using an example with real numbers. Suppose we are given the quadratic <math>x^2-12x+5</math>.<br>, solve for x. | All the different letters are bound to get confusing. It may be easier to visualize this process using an example with real numbers. Suppose we are given the quadratic <math>x^2-12x+5</math>.<br>, solve for x. | ||

The first thing we do is see if it can be factored: it can't. So to solve for x, we will complete the square. | The first thing we do is see if it can be factored: it can't. So to solve for x, we will complete the square. | ||

− | *Step 1 | + | *Step 1 |

+ | {{Hide|1=<math>x^2-12x=-5</math>}} | ||

+ | *Step 2 | ||

+ | {{Hide|The equation will not change from step 1 since the coefficient of the squared term is 1.}} | ||

+ | *Step 3 | ||

+ | {{Hide|1=<math>x^2-12x+36</math>}} | ||

+ | *Step 4 | ||

+ | {{Hide|1=<math>x^2-12x+36=-5+36</math>}} | ||

+ | *Step 5 | ||

+ | {{Hide|1=<math>(x-6)^2-31=0</math>}} |

## Revision as of 10:46, 16 June 2009

## Contents

## The Basics

**Completing the Square** is a method commonly used to solve quadratic equations. Often times, a quadratic equation can be factored and solved easily. However, there are plenty of times when an equation is not factorable. By completing the square, a quadratic equation originally in standard form is rewritten into vertex form.

The equation

is converted into

through the process of **completing the square**.

## Perfect Square Trinomial

A **perfect square trinomial** is a big fancy word for a simple concept. It is a quadratic equation that factors perfectly into two identical binomials.

In general,

The quadratic on the right is a perfect square trinomial. It is the square of a binomial.

#### Example

Take the example of
. Using basic algebra, it can be factored into or

By completing the square, one of the components of the equation has to be a factored perfect square trinomial.

#### Relationship between terms

Note the relationship between the numbers in the factored form and the expanded form. The numbers available are 3, 6, and 9. The middle term of the quadratic is twice the constant of the binomial. The last term of the quadratic is the constant squared. In other words,

and

In general given the perfect trinomial and the equivalent squared binomial

Then,

and

We can rearrange equalities to obtain the following:

## Completing the Square

Let the quadratic equation be

- Step 1: Move the constant over to the other side of the equality

- Step 2: Factor out the coefficient of the squared term

- Step 3: Complete the quadratic in the parenthesis to make a perfect square trinomial. Remember that in general the constant of the perfect square trinomial is the square of half of the coefficient of x.

- Step 4: Add the equivalent added to the left side of the equation, to the right side of the equation to maintain the equality.
*Remember to multiply the obtained constant by a when added to the right side*.

- Step 5: Factor the left side, simplify the right side and bring it over to the left.

- Step 6: If asked for, solve for x.

<template>AlignEquals

|e1l=a(x+\frac{b}{2a})^2+c-\frac{b^2}{4a} |e1r=0 |e2l=a(x+\frac{b}{2a})^2 |e2r=-c+a(\frac{b^2}{4a^2}) |e3l=(x+\frac{b}{2a})^2 |e3r=\frac{-c+a(\frac{b^2}{4a^2})}{a} |e4l=(x+\frac{b}{2a})^2 |e4r=\frac{-c}{a}+\frac{b^2}{4a^2} |e5l=(x+\frac{b}{2a} |e5r </template>

Though the process of completing the square is finalized, it does not look like the vertex form equation

Let <template>AlignEquals

|e1l=-h |e1r=\frac{b}{2a} |e2l=k |e2r=c-\frac{b^2}{4a} </template>

We can now substitute these values into the equation in Step 5 and obtain

### Another Example

All the different letters are bound to get confusing. It may be easier to visualize this process using an example with real numbers. Suppose we are given the quadratic .

, solve for x.
The first thing we do is see if it can be factored: it can't. So to solve for x, we will complete the square.

- Step 1

- Step 2

- Step 3

- Step 4

- Step 5