Difference between revisions of "Completing the Square"

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{{HelperPage|1=Parabola|2=Complex Numbers}}
 
==The Basics==
 
==The Basics==
'''Completing the Square''' is a method used to solve <balloon title="load:quadraticeq">quadratic equations.</balloon> <span id="quadraticeq" style="display:none">A polynomial with highest degree of 2. In general, <math>ax^2+bx+c=0</math></span> When a quadratic is hard to factor or not factorable at all, this method rewrites the quadratic equation originally in <balloon title="load:standardform">standard form</balloon> <span id="standardform" style="display:none"><math>ax^2+bx+c=0</math></span> into <balloon title="load:vertexform">vertex form</balloon> which like a factored quadratic is much easier to solve.  
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'''Completing the Square''' is a method used to solve <balloon title="load:quadraticeq">quadratic equations.</balloon> <span id="quadraticeq" style="display:none">A polynomial with highest degree of 2. In general, <math>ax^2+bx+c=0</math></span> In fact, ''any'' quadratic equation can be solved using this method. When a quadratic is difficult to factor or it involves [[Complex Numbers|complex numbers]], this method rewrites the quadratic equation originally in <balloon title="load:standardform">standard form</balloon> <span id="standardform" style="display:none"><math>ax^2+bx+c=0</math></span> into <balloon title="load:vertexform">vertex form</balloon> which like a factored quadratic is much easier to solve.  
  
 
The equation<br>  
 
The equation<br>  
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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 <balloon title="A polynomial with two terms, for example x+3">binomials.</balloon><br>
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The method of completing the square uses a perfect square trinomial, which allows the quadratic to be easily factored. A '''perfect square trinomial''' 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>
+
where the equation on the left is the expression for all perfect square trinomials.<br>
 
===Example of a Perfect Square Trinomial===
 
===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>
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<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>. The non-factored expression is the perfect square trinomial.
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.<br>
+
<br>
 +
===Obtaining a Perfect Square Trinomial===
 +
A perfect square trinomial can be obtained even when the equation does not immediately factor into one. <br>
 +
Take the example of <math>x^2+10x+14</math> which does not easily factor, nor is it a perfect square trinomial.However, it can be rewritten as the sum of a perfect square trinomial and a constant. <br>
 +
If we wanted the terms <math>x^2+10x</math> to factor into a perfect square such as <math>x^2+2ax+a^2</math>, then we can let <math>10=2a</math>, thus <math>a=5</math> making the last term <math>a^2=25</math>. <br>
 +
As long as the value of the expression does not change terms can be rewritten as combinations of other numbers. This idea is used to obtain a perfect square. We can rewrite 14 as the sum of two numbers, one of which should be 25.<br>
 +
So, <math>x^2+10x+14=x^2+10x+25-11</math><br>
 +
The perfect square trinomial can be factored: <math>(x+5)^2-11</math>
 
====Relationship between terms====
 
====Relationship between terms====
 
*Note that the coefficient of the middle term is twice the square root of the constant term in a quadratic equation.  
 
*Note that the coefficient of the middle term is twice the square root of the constant term in a quadratic equation.  
 
*In other words, the constant term is the square of half of the coefficient of x.
 
*In other words, the constant term is the square of half of the coefficient of x.
*NOTE that this holds only true if the coefficient of <math>x^2</math> is 1.  
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*Note that this holds only true if the coefficient of <math>x^2</math> is 1.  
In general given the perfect trinomial and the equivalent squared binomial <math>x^2+bx+c=(x+d)^2</math><br>
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{{HideThis|1=algebra|2=In general given the perfect trinomial and the equivalent squared binomial <math>x^2+bx+c=(x+d)^2</math><br>
 
Then <math>x^2+bx+c=x^2+2dx+d^2</math><br>
 
Then <math>x^2+bx+c=x^2+2dx+d^2</math><br>
 
So,<br>
 
So,<br>
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and
 
and
 
<math>c=d^2</math><br>
 
<math>c=d^2</math><br>
We can rearrange equalities to obtain the following: <math>c=(\tfrac{b}{2})^2</math>
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We can rearrange equalities to obtain the following: <math>c=\left ( \frac{b}{2}\right ) ^2
 +
 
 
==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 term ''c'' over to the other side of the equal sign.
 
*<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 ''a'', 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 \left( x^2+\frac{b}{a} \right) =-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: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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*<balloon title="load:procedure3">Step 3</balloon> <span id="procedure3" style="display:none"><math>a \left( x^2+\frac{b}{a}+ \left( \frac{b}{2a} \right) ^2 \right)=-c+a \left( \frac{b^2}{4a^2} \right) </math></span>: Complete the quadratic in the parenthesis to make a perfect square trinomial by adding the square of half of the coefficient. Add the equivalent value 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''. <br>
  
*<balloon title="load:procedure5">Step 5</balloon> <span id="procedure5" style="display:none"><math>a(x+\frac{b}{2a})^2+c-\frac{b^2}{4a}=0</math></span>: Factor the left side, simplify the right side and bring it over to the left.<br>
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*<balloon title="load:procedure5">Step 4</balloon> <span id="procedure5" style="display:none"><math>a \left( x+\frac{b}{2a} \right) ^2+c-\frac{b^2}{4a}=0</math></span>: Factor the left side, simplify the right side and bring it over to the left.<br>
*Step 6''(optional)'': If asked for, solve for x.  
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*<balloon title="This step is necessary only if looking for the exact values of x. If the task at hand is to rewrite the equation in vertex form, then this step does not apply.">Step 5</balloon>: If asked for, solve for x.  
 
{{Hide|1=<template>AlignEquals
 
{{Hide|1=<template>AlignEquals
  |e1l=a(x+\frac{b}{2a})^2+c-\frac{b^2}{4a}
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  |e1l=a \left( x+\frac{b}{2a} \right) ^2+c-\frac{b^2}{4a}
 
  |e1r=0
 
  |e1r=0
  |e2l=a(x+\frac{b}{2a})^2
+
  |e2l=a \left( x+\frac{b}{2a} \right) ^2
  |e2r=-c+a(\frac{b^2}{4a^2})
+
  |e2r=-c+a \left( \frac{b^2}{4a^2} \right)
  |e3l=(x+\frac{b}{2a})^2
+
  |e3l= \left( x+\frac{b}{2a} \right) ^2
  |e3r=\frac{-c+a(\frac{b^2}{4a^2})}{a}
+
  |e3r=\frac{-c+a \left( \frac{b^2}{4a^2} \right) }{a}
  |e4l=(x+\frac{b}{2a})^2
+
  |e4l= \left( x+\frac{b}{2a} \right) ^2
 
  |e4r=\frac{-c}{a}+\frac{b^2}{4a^2}
 
  |e4r=\frac{-c}{a}+\frac{b^2}{4a^2}
 
  |e5l=x+\frac{b}{2a}
 
  |e5l=x+\frac{b}{2a}
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  |e6l=x
 
  |e6l=x
 
  |e6r=\pm\sqrt{\frac{-c}{a}+\frac{b^2}{4a^2} }-\frac{b}{2a}
 
  |e6r=\pm\sqrt{\frac{-c}{a}+\frac{b^2}{4a^2} }-\frac{b}{2a}
  </template>}}
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  </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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We can now substitute these values into the equation in Step 5 and obtain <br>
 
We can now substitute these values into the equation in Step 5 and obtain <br>
<math>a(x-h)^2+k=0</math>
+
<math>a(x-h)^2+k=0</math>}}
  
 
===Example with a = 1===
 
===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 <math>x^2-12x+5</math>, and asked to 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=0</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>
 
*<balloon title="load:ex2">Step 2</balloon> <span id="ex2" style="display:none">The equation will not change from step 1 since the coefficient of the squared term is 1.</span>
 
*<balloon title="load:ex2">Step 2</balloon> <span id="ex2" style="display:none">The equation will not change from step 1 since the coefficient of the squared term is 1.</span>
*<balloon title="load:ex3">Step 3</balloon> <span id="ex3" style="display:none"><math>x^2-12x+36</math></span>
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*<balloon title="load:ex3">Step 3</balloon> <span id="ex3" style="display:none"><math>x^2-12x+36=-5+36</math></span>
*<balloon title="load:ex4">Step 4</balloon> <span id="ex4" style="display:none"><math>x^2-12x+36=-5+36</math></span>
+
*<balloon title="load:ex5">Step 4</balloon> <span id="ex5" style="display:none"><math>(x-6)^2-31=0</math></span>
*<balloon title="load:ex5">Step 5</balloon> <span id="ex5" style="display:none"><math>(x-6)^2-31=0</math></span>
+
*<balloon title="load:ex6">Step 5</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>
 
  
===Example with a not 1===
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===Example with <math> a \neq 1 </math>===
Given <math>3x^2+2x-7</math>
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Given <math>3x^2+2x-7=0</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>
*<balloon title="load:examp2">Step 2</balloon> <span id="examp2" style="display:none"><math>3(x^2+\frac{2}{3}x)=7</math></span>
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*<balloon title="load:examp2">Step 2</balloon> <span id="examp2" style="display:none"><math>3 \left( x^2+\frac{2}{3}x \right) =7</math></span>
*<balloon title="load:examp3">Step 3</balloon> <span id="examp3" style="display:none"><math>3(x^2+\frac{2}{3}x+\frac{1}{9}</math></span>
+
*<balloon title="load:examp3">Step 3</balloon> <span id="examp3" style="display:none"><math>3 \left( x^2+\frac{2}{3}x+\frac{1}{9} \right) =7+\frac{1}{3}</math></span>
*<balloon title="load:examp4">Step 4</balloon> <span id="examp4" style="display:none"><math>3(x^2+\frac{2}{3}x+\frac{1}{9}=7+\frac{1}{3}</math></span>
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*<balloon title="load:examp5">Step 4</balloon> <span id="examp5" style="display:none"><math>3 \left( x+\frac{1}{3} \right) ^2-\frac{22}{3}=0</math></span>
*<balloon title="load:examp5">Step 5</balloon> <span id="examp5" style="display:none"><math>3(x+\frac{1}{3})^2-\frac{22}{3}=0</math></span>
+
*<balloon title="load:examp6">Step 5</balloon> <span id="examp6" style="display:none"><math>x=\sqrt{\frac{22}{9} }-\frac{1}{3}</math></span>
*<balloon title="load:examp6">Step 6</balloon> <span id="examp6" style="display:none"><math>x=\sqrt{\frac{22}{9} }-\frac{1}{3}</math></span>
 
  
 
==References==
 
==References==
:*This is not a reference...
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:*http://www.themathpage.com/aPreCalc/complete-the-square.htm
 +
 
 +
:*http://en.wikipedia.org/wiki/Completing_the_square
 +
 
 +
:*http://www.mathsisfun.com/algebra/completing-square.html
 +
 
 +
:*http://www.jamesbrennan.org/algebra/quadratics/Completing%20the%20Square.htm

Latest revision as of 08:44, 16 June 2011

This is a Helper Page for:
Parabola
Complex Numbers

The Basics

Completing the Square is a method used to solve quadratic equations. In fact, any quadratic equation can be solved using this method. When a quadratic is difficult to factor or it involves complex numbers, 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

The method of completing the square uses a perfect square trinomial, which allows the quadratic to be easily factored. A perfect square trinomial 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
where the equation on the left is the expression for all perfect square trinomials.

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. The non-factored expression is the perfect square trinomial.

Obtaining a Perfect Square Trinomial

A perfect square trinomial can be obtained even when the equation does not immediately factor into one.
Take the example of x^2+10x+14 which does not easily factor, nor is it a perfect square trinomial.However, it can be rewritten as the sum of a perfect square trinomial and a constant.
If we wanted the terms x^2+10x to factor into a perfect square such as x^2+2ax+a^2, then we can let 10=2a, thus a=5 making the last term a^2=25.
As long as the value of the expression does not change terms can be rewritten as combinations of other numbers. This idea is used to obtain a perfect square. We can rewrite 14 as the sum of two numbers, one of which should be 25.
So, x^2+10x+14=x^2+10x+25-11
The perfect square trinomial can be factored: (x+5)^2-11

Relationship between terms

  • Note that the coefficient of the middle term is twice the square root of the constant term in a quadratic equation.
  • In other words, the constant term is the square of half of the coefficient of x.
  • Note that this holds only true if the coefficient of x^2 is 1.

{{HideThis|1=algebra|2=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=\left ( \frac{b}{2}\right ) ^2

==Procedure for Completing the Square==
Let the quadratic equation be <math>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 by adding the square of half of the coefficient. Add the equivalent value 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 : Factor the left side, simplify the right side and bring it over to the left.
  • Step 5: If asked for, solve for x.

<template>AlignEquals

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=0, 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 \neq 1

Given 3x^2+2x-7=0

References