Solver91311
Deriving the Quadratic Formula
and Evaluating the Discriminant
Where did that crazy formula come from anyway?
John Reece
7/20/2010
Let’s review the Complete the Square process
Remember that there is a seldom used process for solving quadratic equations
called Completing the Square. I think that we only learn it so that we can see how the
Quadratic Formula is derived. Be that as it may, let’s look at the process.
Example 1: x 2  x  1  0
Step 1: Move the constant term to the RHS
x2  x  1
Step 2: Divide through by the lead coefficient. Lead coefficient on this one is 1, so
skip this step.
Step 3: Divide the coefficient on the 1st degree term by 2, square the result, then add
that result to both sides:
x2  x 
1 5

4 4
Step 4: Factor the Perfect Square LHS
 x  12 
2

5
4
Step 5: Take the square root of both sides:
5
x 1 
2
2
Step 6: Add the opposite of the constant in the LHS to both sides:
x
1 5
2
Example 2: 3x 2  5 x  15  0
Step 1: Move the constant term to the RHS
3x 2  5 x  15
Step 2: Divide through by the lead coefficient.
5
x2  x  5
3
Step 3: Divide the coefficient on the 1st degree term by 2, square the result, then add
that result to both sides:
5
25
25 205
x2  x 
 5

3
36
36 36
Step 4: Factor the Perfect Square LHS
2
5
205

x  
6
36

Step 5: Take the square root of both sides:
x
5
205

6
6
Step 6: Add the opposite of the constant in the LHS to both sides:
x
5  205
6
Deriving the Quadratic Formula
Just like the two examples above, but instead of specific values for the coefficients,
we’re going to us the general quadratic equation:
ax 2  bx  c  0
Step 1: Move the constant term to the RHS
ax 2  bx  c
Step 2: Divide through by the lead coefficient.
x2 
b
c
x
a
a
Step 3: Divide the coefficient on the 1st degree term by 2, square the result, then add
that result to both sides:
x2 
b
b2
b 2  4ac
x 2 
a
4a
4a 2
Step 4: Factor the Perfect Square LHS
b  b 2  4ac

x


 
2a 
4a 2

2
Step 5: Take the square root of both sides:
x
b
b 2  4ac

2a
2a
Step 6: Add the opposite of the constant in the LHS to both sides:
x
b  b 2  4ac
2a
which you should recognize as the general solution for a quadratic equation in
standard form.
Evaluating the Discriminant
The discriminant is a handy tool to use to determine the character of the roots of a
quadratic equation. Recall that there are three possibilities: 1. The roots are real and
equal (alternatively, you can say there is one root with a multiplicity of two.) 2. The
roots are real and unequal. 3. The roots are a conjugate pair of complex numbers of
the form    i
Let’s first recall the quadratic formula: x 
b  b 2  4ac
. The discriminant is nothing
2a
more than the part of the quadratic formula that is inside of the radical, namely:
  b 2  4ac
Here’s what the discriminant tells us:
If   0 , then there are two real and equal roots, or one root with a multiplicity of two if
you prefer. Visually speaking, the graph of the function that corresponds to the
quadratic equation is tangent to the x-axis at the vertex of the parabola.
If   0 , then there are two real and unequal roots. Additionally, if  is a perfect
square, then the two roots are rational, otherwise not. Visually speaking, the graph of
the function that corresponds to the quadratic equation intercepts the x-axis at two
distinct points.
If   0 , then there is a conjugate pair of complex roots of the form    i where
 ,    and i is the imaginary number defined by i 2  1 . Visually speaking, the
graph of the function that corresponds to the quadratic equation does not intercept
the x-axis.