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Linear - Quadratic Systems
Topic Index | Algebra2/Trig Index | Regents Exam Prep Center
A quadratic equation is defined as an equation in which one or more of the terms is squared but raise
to no higher power. The general form is ax2 + bx + c = 0, where a, b and c are constants.
In Algebra and Geometry, we learned how to solve linear - quadratic systems algebraically and
graphically. With our new found knowledge of quadratics, we are now ready to attack problems tha
cannot be solved by factoring, and problems with no real solutions.
The familiar linear-quadratic system:
(where the quadratic is in one variable)
Remember that linear-quadratic systems of this type can result in three graphical situations such as
The equations will intersect in two
locations. Two real solutions.
The equations will intersect in one
location. One real solution.
The equations will not intersect.
No real solutions.
Keep these images in mind as we proceed to solve these linear-quadratic systems algebraically.
Example 1:
When we studied these systems in Algebra, we encountered situations that could be solved by
factoring, such as this first example.
Solve this system of equations algebraically:
y = x2 - x - 6 (quadratic equation in one variable of form y = ax2 + bx + c )
y = 2x - 2
(linear equation of form y = mx + b)
Substitute from the linear
equation into the quadratic
equation and solve.
y = x2 - x - 6
2x - 2 = x2 - x - 6
2x = x2 - x - 4
0 = x2 - 3x - 4
Find the y-values by substituting There are 2 "possible" solutions
each value of x into the linear for the system: (4,6) and (-1,-4)
equation.
Check each in both equations.
y = 2(4) - 2 = 6
POINT (4,6)
y = 2(-1) - 2 = -4
POINT (-1,-4)
0 =(x - 4)(x + 1)
x-4=0
x=4
x + 1 =0
x = -1
See how to use your
TI-83+/84+ graphing
calculator with
quadratic-linear
systems.
Click calculator.
y = x2 - x - 6
6 = (4)2 - 4 - 6 = 6 checks
y = 2x - 2
6 = 2(4) - 2 = 6 checks
y = x2 - x - 6
-4 = (-1)2 - (-1) - 6 = -4 checks
y = 2x - 2
-4 = 2(-1) - 2 = -4 checks
Answer:
{(4, 6), (-1, -4)}
Example 2:
With our new found knowledge of quadratics, we are now ready to attack problems that cannot
be solved by factoring, and/or problems with no real solutions (such as this second example).
Solve this system of equations algebraically:
y = x2 - 2x + 1 (quadratic equation in one variable of form y = ax2 + bx + c )
y=x-3
(linear equation of form y = mx + b)
Substitute from the linear
equation into the quadratic
equation.
y = x2 - 2x + 1
x - 3 = x2 - 2x + 1
0 = x2 - 3x + 4
Use quadratic formula:
Find the y-values
by substituting each value
of xinto the linear
equation.
y = x2 - 2x + 1
POINT
No real solutions.
There are 2 "possible"
solutions.Check each in both
equations.
y=x-3
POINT
Answer:
There are no real solutions.
The answers are complex numbers,
which are not graphed in the Cartesian coordinate plane.
------------------------------------y = x2 - 2x + 1
y=x-3
Example 1:
Solve this system of equations algebraically:
y = x2 - x - 6 (quadratic equation of
form y = ax2 + bx + c: parabola)
y = 2x - 2
(linear equation of form y = mx + b)
Since this is already done for us in
this example, we can go to the
next step.
First, we solve for one
of the variables in the
linear equation.
y = 2x - 2
Next, we substitute for
that variable in the
quadratic equation, and
solve the resulting
equation.
y = x2 - x - 6
2x - 2 = x2 - x - 6
2x = x2 - x - 4
Add 2 to both sides.
Subtract 2x from both sides.
0 = x2 - 3x - 4
Factor.
0 =(x - 4)(x + 1)
x-4=0
x=4
x + 1 =0
x = -1
Set each factor = 0 and solve.
We now have two values for x, but we still need to find the corresponding values for y.
We find the y-values by
substituting each value
of x into the linear
equation.
y = 2x - 2
Now we have 2 possible
solutions for the system: (4,6)
and (-1,-4). We need
tocheck each solution in each
equation.
Check#1: (4, 6)
Check#2: (-1, -4)
y = x2 - x - 6
y = x2 - x - 6
6 = (4)2 - 4 - 6
6 = 16 - 4 - 6
6 = 6 it checks !
-4 = (-1)2 - (-1) - 6
-4 = 1 + 1 - 6
-4 = -4 it checks !
y = 2x - 2
y = 2x - 2
6 = 2(4) - 2
6=8-2
6 = 6 it also checks !
-4 = 2(-1) - 2
-4 = -2 - 2
-4 = -4 it also checks !
y = 2(4) - 2
y=8-2
y=6
Check 4
(4, 6)
y = 2x - 2
y = 2(-1) - 2
y = -2 - 2
y = -4
Check -1
(-1, -4)
We finally have our solution
set for this linear quadratic
system.
{(4, 6), (-1, -4)}