Now we have a quadratic equation in standard form. That means we can solve it directly with the
quadratic formula. Let’s state the quadratic formula once again as we originally learned it. For the general
quadratic
ax^2 +bx+c= 0
the roots are given by
x= [−b± (b^2 − 4 ac)1/2] / (2a)
To make the quadratic formula work in the current equation, let’s make these substitutions in the classical
version:
- Write a 2 in place of a
- Write (b 2 −a 1 ) in place of b
- Write (c−b 1 ) in place of c
When we do that, we get
x= {−(b 2 −a 1 )± [(b 2 −a 1 )^2 − 4 a 2 (c−b 1 )]1/2} / (2a 2 )
The roots defined by this formula give us the x-values for the solution of the original linear-quadratic sys-
tem. We can simplify this slightly by getting rid of the minus sign in the first expression on the right side
of the equals sign, and then reversing the positions of a 1 and b 2 inside the parentheses, getting
x= {(a 1 −b 2 )± [(b 2 −a 1 )^2 − 4 a 2 (c−b 1 )]1/2} / (2a 2 )
Any attempt to further simplify this will eliminate some grouping symbols, but will not make the formula
easier to use.
When we’ve found the x-values of the solutions, we can plug them into either of the original functions
to obtain the y-values. In this case, the linear function is less messy than the quadratic. If we call the x-values
x 1 and x 2 , then
y 1 =a 1 x 1 +b 1
and
y 2 =a 1 x 2 +b 1
The solutions of the whole system can be written as the ordered pairs (x 1 ,y 1 ) and (x 2 ,y 2 ).
Two Quadratics
Now let’s solve a two-by-two system consisting of these quadratic equations in the two vari-
ablesx and y:
4 x^2 + 6 x+ 2 y+ 8 = 0
Two Quadratics 451
