Difference between revisions of "Completing the Square"

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==Perfect Square Trinomial==
 
==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.<br>
+
A '''perfect square trinomial''' is a big fancy word for a simple concept. It is a quadratic equation that factors perfectly into two identical <balloon title="A polynomial with two terms, for example x+3">binomials.</balloon><br>
 
In general, <br>
 
In general, <br>
 
<math>x^2-2ax+a^2=(x-a)(x-a)=(x-a)^2</math><br>
 
<math>x^2-2ax+a^2=(x-a)(x-a)=(x-a)^2</math><br>
 
The quadratic on the left is a perfect square trinomial. It is the square of a binomial.<br>
 
The quadratic on the left is a perfect square trinomial. It is the square of a binomial.<br>
====Example====
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===Example of a Perfect Square Trinomial===
 
Take the example of  
 
Take the example of  
 
<math>x^2+6x+9</math>. It can be factored into <math>(x+3)(x+3)</math> whcih is equal to <math>(x+3)^2</math><br>
 
<math>x^2+6x+9</math>. It can be factored into <math>(x+3)(x+3)</math> whcih is equal to <math>(x+3)^2</math><br>
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<math>2\times3=6</math> and <math>3\times3=9</math><br>
 
<math>2\times3=6</math> and <math>3\times3=9</math><br>
 
In general given the perfect trinomial and the equivalent squared binomial <math>x^2+bx+c=(x+d)^2</math><br>
 
In general given the perfect trinomial and the equivalent squared binomial <math>x^2+bx+c=(x+d)^2</math><br>
Then,<br>
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Then <math>x^2+bx+c=x^2+2dx+d^2</math><br>
 +
So,<br>
 
<math>b=2d</math>
 
<math>b=2d</math>
 
and
 
and
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==Procedure for Completing the Square==
 
==Procedure for Completing the Square==
 
Let the quadratic equation be <math>ax^2+bx+c=0</math>
 
Let the quadratic equation be <math>ax^2+bx+c=0</math>
*<balloon title="load:procedure1">Step 1</balloon> <span id="procedure1" style="display:none"><math>ax^2+bx=-c</math></span>: Move the constant over to the other side of the equality
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*<balloon title="load:procedure1">Step 1</balloon> <span id="procedure1" style="display:none"><math>ax^2+bx=-c</math></span>: Move the constant term ''c'' over to the other side of the equal sign.
*<balloon title="load:procedure2">Step 2</balloon> <span id="procedure2" style="display:none"><math>a(x^2+\frac{b}{a})=-c</math></span>: Factor out the coefficient of the squared term <br>
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*<balloon title="load:procedure2">Step 2</balloon> <span id="procedure2" style="display:none"><math>a(x^2+\frac{b}{a})=-c</math></span>: Factor out ''a'', the coefficient of the squared term. <br>
 
*<balloon title="load:procedure3">Step 3</balloon> <span id="procedure3" style="display:none"><math>a(x^2+\frac{b}{a}+(\frac{b}{2a})^2)=a(x^2+\frac{b}{a}+\frac{b^2}{4a^2})</math></span>: 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>
 
*<balloon title="load:procedure3">Step 3</balloon> <span id="procedure3" style="display:none"><math>a(x^2+\frac{b}{a}+(\frac{b}{2a})^2)=a(x^2+\frac{b}{a}+\frac{b^2}{4a^2})</math></span>: 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>
 
*<balloon title="load:procedure4">Step 4</balloon> <span id="procedure4" style="display:none"><math>a(x^2+\frac{b}{a}+\frac{b^2}{4a^2})=-c+a(\frac{b^2}{4a^2})</math></span>: 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''.
 
*<balloon title="load:procedure4">Step 4</balloon> <span id="procedure4" style="display:none"><math>a(x^2+\frac{b}{a}+\frac{b^2}{4a^2})=-c+a(\frac{b^2}{4a^2})</math></span>: 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''.
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<math>a(x-h)^2+k=0</math>
 
<math>a(x-h)^2+k=0</math>
  
===Another Example===
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===Example with a = 1===
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 understand the process using an example with real numbers. Suppose we are given the quadratic <math>x^2-12x+5</math>, and asked to 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.
 
*<balloon title="load:ex1">Step 1</balloon> <span id="ex1" style="display:none"><math>x^2-12x=-5</math></span>
 
*<balloon title="load:ex1">Step 1</balloon> <span id="ex1" style="display:none"><math>x^2-12x=-5</math></span>
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*<balloon title="load:ex6">Step 6</balloon> <span id="ex6" style="display:none"><math>x=\pm\sqrt{31}+6</math></span>
 
*<balloon title="load:ex6">Step 6</balloon> <span id="ex6" style="display:none"><math>x=\pm\sqrt{31}+6</math></span>
  
===One Last Example===
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===Example with a not 1===
 
Given <math>3x^2+2x-7</math>
 
Given <math>3x^2+2x-7</math>
 
*<balloon title="load:examp1">Step 1</balloon> <span id="examp1" style="display:none"><math>3x^2+2x=7</math></span>
 
*<balloon title="load:examp1">Step 1</balloon> <span id="examp1" style="display:none"><math>3x^2+2x=7</math></span>

Revision as of 11:10, 18 June 2009

The Basics

Completing the Square is a method used to solve quadratic equations. When a quadratic is hard to factor or not factorable at all, this method rewrites the quadratic equation originally in standard form into vertex form which like a factored quadratic is much easier to solve.

The equation
 ax^2+bx+c=0
is converted into
a(x-h)^2+k=0

and x=\pm\sqrt{\frac{-k}{a}}+h

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,
x^2-2ax+a^2=(x-a)(x-a)=(x-a)^2
The quadratic on the left is a perfect square trinomial. It is the square of a binomial.

Example of a Perfect Square Trinomial

Take the example of x^2+6x+9. It can be factored into (x+3)(x+3) whcih is equal to (x+3)^2
This concept is important because in the process of 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,
2\times3=6 and 3\times3=9
In general given the perfect trinomial and the equivalent squared binomial x^2+bx+c=(x+d)^2
Then x^2+bx+c=x^2+2dx+d^2
So,
b=2d and c=d^2
We can rearrange equalities to obtain the following: c=(\tfrac{b}{2})^2

Procedure for Completing the Square

Let the quadratic equation be ax^2+bx+c=0

  • Step 1 : Move the constant term c over to the other side of the equal sign.
  • Step 2 : Factor out a, 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(optional): If asked for, solve for x.

<template>AlignEquals

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
a(x-h)^2+k=0

Example with a = 1

All the different letters are bound to get confusing. It may be easier to understand the process using an example with real numbers. Suppose we are given the quadratic x^2-12x+5, and asked to 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.

Example with a not 1

Given 3x^2+2x-7