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Difference between revisions of "Completing the square"

(Nice article, should we add an example?)
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===Motivations===
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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.
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Quadratic equations are usally hard to solve, if they do not factor.
 
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Some of them, in forms <math>(x+a)^2=b</math>, are easy to solve by taking the squareroot of b and subracting a. Completing the square is a technique to munipulate ''every'' quadratic into the easily solve-able form above.
 
  
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===General Solution For A Monic Quadratic===
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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
  
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Let the quadratic be in the form <math>x^2+xa+b=0</math>.
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<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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Moving be to the other side, we obtain
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==Motivations==
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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.
  
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<math>x^2+xa=-b</math>
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==General Solution For A Quadratic by Completing the Square==
  
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Then, adding a^2/4 to each side and factoring, we get
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Let the quadratic be in the form <math>a\cdot x^2+b\cdot x+c=0</math>, with <math>a \neq 0</math>.
  
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<math>(x+1/2a)^2=-b+a^2/4</math>
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Subtracting <math>c</math> from both sides of the equation, we obtain
  
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Which is solvable as described above.
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<math>ax^2+bx=-c</math>.
  
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===Applications of Adding and Factoring===
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Dividing by <math>{a}</math> and adding <math>\frac{b^2}{4a^2}</math> to both sides yields
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<math>x^2+\frac{b}{a}x+\frac{b^2}{4a^2}=-\frac{c}{a}+\frac{b^2}{4a^2}</math>.
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Factoring the LHS gives
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<math>\left(x+\frac{b}{2a}\right)^2=\frac{b^2-4ac}{4a^2}</math>
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As described above, an equation in this form can be solved, yielding
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<math>{x=\frac{-b \pm \sqrt{b^2-4ac}}{2a}}</math>.
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This formula is also called the [[quadratic formula]].
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==Applications of Adding and Factoring==
 
Other degrees of polynomials may be solved by adding constant terms 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 <math>(x-3)^2</math>. 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.
 
Another common usage is in conic sections. The equations for conic sections typically contain a squared term such as <math>(x-3)^2</math>. 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.
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All kinds of exotic factoring techniques are used on the [[AIME]], including completing the square.
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==Problems==
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===Introductory===
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<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===
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*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===
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[[Category:Algebra]]
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[[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 $x^2-6x+2=0$. Look at the $x^2-6x$ part. If $9$ was added to this, then we would have a perfect square, $(x-3)^2=x^2-6x+9$. To do this, add $7$ to each side of the equation to get

$x^2-6x+9=7\Rightarrow(x-3)^2=7\Rightarrow x-3=\pm{\sqrt7}\Rightarrow x=3\pm\sqrt{7}$

Motivations

All quadratic equations in the form $(x+a)^2=b$ 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.

General Solution For A Quadratic by Completing the Square

Let the quadratic be in the form $a\cdot x^2+b\cdot x+c=0$, with $a \neq 0$.

Subtracting $c$ from both sides of the equation, we obtain

$ax^2+bx=-c$.

Dividing by ${a}$ and adding $\frac{b^2}{4a^2}$ to both sides yields

$x^2+\frac{b}{a}x+\frac{b^2}{4a^2}=-\frac{c}{a}+\frac{b^2}{4a^2}$.

Factoring the LHS gives

$\left(x+\frac{b}{2a}\right)^2=\frac{b^2-4ac}{4a^2}$

As described above, an equation in this form can be solved, yielding

${x=\frac{-b \pm \sqrt{b^2-4ac}}{2a}}$.

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 $(x-3)^2$. 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

$x^2+2x=28$ solve for x

What is the value of $x$ if $x=1+\dfrac{1}{x}$

Intermediate

Olympiad

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