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Difference between revisions of "Completing the square"

(→‎General Solution For A Quadratic by Completing the Square)
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===General Solution For A Quadratic by Completing the Square===
 
===General Solution For A Quadratic by Completing the Square===
  
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Let the quadratic be in the form <math>ax^2+bx+c=0</math>.  
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Let the quadratic be in the form <math>a*x^2+b*x+c=0</math>.  
  
 
Moving c to the other side, we obtain
 
Moving c to the other side, we obtain

Revision as of 17:57, 17 June 2006

Motivations

All quadratic equations in the form $(x+a)^2=b$ can be solved by taking the square root of b and subracting a. Completing the square is a technique to manipulate every quadratic into the easily solveable form above.

General Solution For A Quadratic by Completing the Square

Let the quadratic be in the form $a*x^2+b*x+c=0$.

Moving c to the other side, we obtain

$a*x^2+b*x=-c$

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, and completing the square is just one. The follow solution to AIME 1987 #5 is but one.

Instructive Problem: AIME 1987 #5

Problem:m, n are integers such that $m^2 + 3m^2n^2 = 30n^2 + 517$. Find $3m^2n^2$.

Outline Solution: We use the techinique of adding to factor.

Rearrange to $m^2 + 3m^2n^2 -30n^2= 517$. The key step is changing the equation to $m^2 + 3m^2n^2 -30n^2-10= 507$, where the equation factors to $(3n^2 + 1)(m^2 - 10) = 507 = 3*13^2$, from which the problem is trivial to solve by applying some simple number theory.

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