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 9 was added to this, then we would have a [[perfect square]], <math>(x-3)^2=x^2-6x+9</math>.  To do this, add 7 to each side of the equation to get  
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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  
  
 
<math>x^2-6x+9=7\Rightarrow(x-3)^2=7\Rightarrow x-3=\pm{\sqrt7}\Rightarrow x=3\pm\sqrt{7}</math>
 
<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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==Problems==
 
==Problems==
 
===Introductory===
 
===Introductory===
<math>x^2+2x=28 solve for x</math>
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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>
  
 
===Intermediate===
 
===Intermediate===
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===Olympiad===
 
===Olympiad===
  
[[Category:Elementary algebra]]
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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