Difference between revisions of "Completing the square"
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| − | + | The idea of '''completing the square''' is to add something to an equation to make that equation a [[perfect square]]. This makes solving a lot of equations easy. In fact, all [[quadratic equation]]s can be solved by completing the square. | |
| − | The idea of completing the square is to add something to an equation to make that equation a [[perfect square]]. This makes solving a lot of equations easy. In fact, all [[quadratic equation]]s can be solved by completing the square. | ||
| + | As an example, we have the equation <math>x^2-6x+2=0</math>. Look at the <math>x^2-6x</math> part. If <math>9</math> was added to this, then we would have a [[perfect square]], <math>(x-3)^2=x^2-6x+9</math>. To do this, add <math>7</math> to each side of the equation to get | ||
| − | + | <math>x^2-6x+9=7\Rightarrow(x-3)^2=7\Rightarrow x-3=\pm{\sqrt7}\Rightarrow x=3\pm\sqrt{7}</math> | |
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==Motivations== | ==Motivations== | ||
| − | All quadratic equations in the form <math>(x+a)^2=b</math> can be solved by taking the square root of b and subtracting a. Completing the square is a technique to manipulate ''every'' quadratic into the easily solvable form above. | + | All quadratic equations in the form <math>(x+a)^2=b</math> can be solved by taking the [[square root]] of <math>b</math> and subtracting <math>a</math>. Completing the square is a technique to manipulate ''every'' quadratic into the easily solvable form above. |
==General Solution For A Quadratic by Completing the Square== | ==General Solution For A Quadratic by Completing the Square== | ||
| − | Let the quadratic be in the form <math>a\cdot x^2+b\cdot x+c=0</math>. | + | Let the quadratic be in the form <math>a\cdot x^2+b\cdot x+c=0</math>, with <math>a \neq 0</math>. |
| − | + | Subtracting <math>c</math> from both sides of the equation, we obtain | |
| − | <math>ax^2+bx=-c</math> | + | <math>ax^2+bx=-c</math>. |
Dividing by <math>{a}</math> and adding <math>\frac{b^2}{4a^2}</math> to both sides yields | Dividing by <math>{a}</math> and adding <math>\frac{b^2}{4a^2}</math> to both sides yields | ||
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As described above, an equation in this form can be solved, yielding | As described above, an equation in this form can be solved, yielding | ||
| − | <math>{x=\frac{-b \pm \sqrt{b^2-4ac}}{2a}}</math> | + | <math>{x=\frac{-b \pm \sqrt{b^2-4ac}}{2a}}</math>. |
| − | This formula is also called the [[ | + | This formula is also called the [[quadratic formula]]. |
==Applications of Adding and Factoring== | ==Applications of Adding and Factoring== | ||
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==Problems== | ==Problems== | ||
===Introductory=== | ===Introductory=== | ||
| + | <math>x^2+2x=28 </math> solve for x | ||
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| + | What is the value of <math>x</math> if <math>x=1+\dfrac{1}{x}</math> | ||
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===Intermediate=== | ===Intermediate=== | ||
*Find <math>3x^2 y^2</math> if <math>x</math> and <math>y</math> are [[integer]]s such that <math>y^2 + 3x^2 y^2 = 30x^2 + 517</math>. ([[1987 AIME Problems/Problem 5|Source]]) | *Find <math>3x^2 y^2</math> if <math>x</math> and <math>y</math> are [[integer]]s such that <math>y^2 + 3x^2 y^2 = 30x^2 + 517</math>. ([[1987 AIME Problems/Problem 5|Source]]) | ||
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===Olympiad=== | ===Olympiad=== | ||
| − | [[Category: | + | [[Category:Algebra]] |
| + | [[Category:Quadratic equations]] | ||
Latest revision as of 15:20, 5 March 2023
The idea of completing the square is to add something to an equation to make that equation a perfect square. This makes solving a lot of equations easy. In fact, all quadratic equations can be solved by completing the square.
As an example, we have the equation 
. Look at the 
part. If 
was added to this, then we would have a perfect square, 
. To do this, add 
to each side of the equation to get
Contents
Motivations
All quadratic equations in the form 
can be solved by taking the square root of 
and subtracting 
. Completing the square is a technique to manipulate every quadratic into the easily solvable form above.
General Solution For A Quadratic by Completing the Square
Let the quadratic be in the form 
, with 
.
Subtracting 
from both sides of the equation, we obtain
.
Dividing by 
and adding 
to both sides yields
.
Factoring the LHS gives
As described above, an equation in this form can be solved, yielding
.
This formula is also called the quadratic formula.
Applications of Adding and Factoring
Other degrees of polynomials may be solved by adding constant terms and factoring.
Another common usage is in conic sections. The equations for conic sections typically contain a squared term such as 
. However, the problem may be posed as to convert from an expanded form to a factored perfect square. Completing the square is the standard method.
All kinds of exotic factoring techniques are used on the AIME, including completing the square.
Problems
Introductory
solve for x
What is the value of 
if 
if 
are 
. (
