Completing the Square
and the Quadratic Formula
In principle, any polynomial can be solved by factoring.  However even  quadratic polynomials can be very difficult to factor using just the informal approach we took in the introduction to factoring.  We will now present two closely related methods for solving any quadratic polynomial:
  1. "Completing the Square"
  2. The quadratic formula.
"Completing the square" requires a small amount of thought to carry through, while the quadratic formula provides a means for solving quadratics by just "turning the crank".  We will explain "completing the square" by analyzing it geometrically.   We will also show how the quadratic formula can be derived using "completing the square"

 A definition:
A polynomial of the form x2 +2ax + a2 is the square of the monomial x + a, that is  x2 +2ax + a2 = (x +a)2 .  Such polynomials are called squares for the simple reason that for any choice of a and x a square with sides of length  a+x  has area  x2 +2ax + a2.
x = -2 6

An example:
Consider the polynomial x2 +4x - 2 .   The factors of this polynomial are  (x + 2 +6) and (x + 2 - 6) and so the solutions are  x = -26.  It is easy to check that this answer is correct.  However, the real problem  is to find the factors or solutions in the first place. Here's a way we could do it:

  1. What we want to do is solve the equation x2 +4x - 2= 0.
  2. If we add 6 to both sides then we  get the equation  x2 +4x +4= 6
  3. Observe that the left side of this equation is a square, that is  x2 +4x +4 = (x+2)2
  4. If  we take the square root of both sides of the quadratic equation x2 +4x +4= 6  we get the linear equations x + 2 = 6  (because of the use  of  this is actually two equations: x + 2 = 6 and x + 2 = -6 ).
  5. Subtracting 2 from both sides of these equations  we get x = -2 6, that is x = -2 +6 or x = -2 -6
The ideas behind "completing the square":
  1. An equation of the form  x2 +2ax + a2 = n, where the left side is a square and the right side is a number, is very easy to solve.  All we have to do is take square root of both sides, getting  x + a = n, and then solve this linear equation to get x = -a n as the solutions.
  2. Solving a polynomial x2 +bx + c  is the same as solving the equation x2 +bx + c = 0, but if we add  -c +( b/2)2 to both sides we get the equation x2 +bx + (b/2)2 = -c+(b/2)2  which is of  the  form x2 +2ax + a2 = n if we take a = b/2 and n = - c +(b/2)2 .  This simple transformation allows us to use the special method for solving equations where the left side is a square and the right side is a number.

    3.  
Here's an algorithm for "completing the square":   Given a polynomial x2 +bx + c  that you wish to solve, proceed as follows:
  1. Rewrite the polynomial as the equation  x2 +bx  =  - c.
  2. Add  (b/2)2 to both sides getting   the new equation x2 +bx + (b/2)2 =  - c+ (b/2)2
  3. Note that  x2 +bx + (b/2)2 is a square, that is: x2 +bx + (b/2)2 =  (x + (b/2))2
  4. Take the square-root of both sides of the equation x2 +bx + (b/2)2 =  - c+ (b/2)2  getting the new equation x + b/2 = (-c + (b/2)2 )
  5. Solve that equation for x  getting x  = - b/2  (-c + (b/2)2 )

    6.  
The same algorithm can be used to solve polynomials of the form ax2 +bx + c. All you have to do is first divide the polynomial  ax2 +bx + c by a to get the polynomial  x2 +(b/a)x +( c/a)  which is of the correct form for the application of the algorithm.   Question for the student:  what rule justifies this division -- that is, what rule tells us that polynomials  ax2 +bx + c and  x2 +(b/a)x +( c/a) have the same solutions?

Carrying through the algorithm starting from the polynomial  ax2 +bx + c (or x2 +(b/a)x +( c/a) )  gives us the result that the solutions are given by the formula

This formula is called the quadratic formula.  This formula can be used instead of the  algorithm for completing the square.   All you have to do is put in the appropriate values for the coeficients of the polynomial and then evaluate the resulting expression.

The subformula, ( b2 - 4ac) is called the discriminant.  It can be used to get information about the nature of the solutions of the particular quadratic.