Evaluating the Quadratic Formula
Solving a Quadratic Equation Graphically
From a graphical point of view, the
solutions of a quadratic equation like:
3Â· x^{ 2} + 6 Â· x +1= 0
are the xcoordinates of the points
where the graph o f y = 3Â· x^{ 2} + 6 Â· x +1 cuts through the
xaxis
(see Figure 1).
Figure 1: The solutions of a quadratic are the xcoordinates
of the points where the graph cuts the
xaxis.
You can locate these points (at least
approximately) by using a graphing
calculator.
Solving a Quadratic Equation
Algebraically
The key difficulty in using a graphing
calculator to solve a quadratic
equation (in fact, the main problem in
using a graphing calculator to solve
any equation) is that you need to set
the viewing window so that you can
see the points where the quadratic graph cuts through the xaxis. Unless you happen to
know about where these points will be, setting the viewing window to show them is a hitand
miss, trialanderror procedure.
As with many of the other equations that we have studied, it is possible to solve a
quadratic equation using algebra. Using algebra to solve a quadratic equation avoids the
problem of guessing a good size for the viewing window and is often easier because of
this.
Using Factoring to Solve a Quadratic Equation
Unless the numbers are chosen to be very nice, factoring a quadratic equation can be very
hard or impossible. However, when you are able to factor a quadratic, the factored form
makes it very easy for you to find the solutions of the quadratic equation.
Example
Find all solutions of the quadratic equation:
x^{ 2} + 8 Â· x +10 = 30.
Solution
First note that solving this quadratic equation is the same as solving the quadratic
equation:
x^{ 2} + 8 Â· x +10  30 = 30  30 (Subtract 30 from each side)
x^{ 2} + 8 Â· x  20 = 0 (Simplify)
Solving the quadratic equation x^{2} + 8 Â· x  20 = 0 will give exactly the same values for x
that solving the original quadratic equation, x^{2} + 8 Â· x +10 = 30, will give.
To factor the quadratic, you have to find a pair of numbers that add to give +8 and which
multiply to give 20. Two numbers that do this are +10 and 2. So you can write the
quadratic equation in factored form as:
(x  2) Â· (x +10) = 0
The two xvalues that will satisfy this equation are x = 2 and x = 10. The solutions of
the quadratic equation:
x^{2} + 8 Â· x +10 = 30
are x = 2 and x = 10.
Using Vertex Form to Solve a Quadratic Equation
If you happen to have a quadratic equation where the quadratic part of the equation is
already expressed in vertex form, then you can solve the quadratic equation in a
reasonably straightforward manner.
If the quadratic part of your quadratic equation is not already expressed in vertex form,
then you can convert it to vertex form by completing the square. Completing the square
can be quite a complicated process, however, so solving a quadratic equation that is not
already in vertex form can be quite an undertaking.
