Linear - Quadratic Systems
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A quadratic equation is defined as an equation in which one or more of the terms is squared but raised 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 that 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

0 =(x - 4)(x + 1)

x - 4 = 0      x + 1 =0
x = 4           x = -1

Find the y-values by substituting each value of x into the linear  
equation.


 
y = 2(4) - 2 = 6
POINT (4,6)

y = 2(-1) - 2 = -4
POINT (-1,-4)

See how to use your
TI-83+/84+ graphing calculator  with quadratic-linear systems.
Click calculator.

There are 2 "possible" solutions for the system:  (4,6) and (-1,-4)
Check each in both equations.

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:


No real solutions.

Find the y-values by substituting each value of x into the linear  
equation.


 

POINT

POINT

There are 2 "possible" solutions. Check each in both equations.

y = x2 - 2x + 1

y = x - 3

  -------------------------------------
y = x2 - 2x + 1


y = x - 3

Answer:
There are no real solutions.
The answers are complex numbers,
which are not graphed in the Cartesian coordinate plane.
 

 

 

Other linear - quadratic systems:
(where the quadratic is in two variables)

Quadratics in two variables look like x2 + y2 = 16 where two variables are squared.
 

Example 3:

Solve this system of equations algebraically:
                 x2 + y2 = 25 
(quadratic equation of a circle center (0,0), radius 5)
                 4y = 3x     
    (linear equation)

Substitute from the linear equation into the quadratic equation and solve.

Find the y-values by substituting each value of x into the linear  
equation.


  

Answer:

{(4, 3), (-4, -3)}

There are 2 "possible" solutions for the system:  (4,3) and (-4,-3)
Check each in both equations.

x2 + y2 = 25
42 + 32 = 25
1
6 + 9 = 25
25 = 25
checks
4y = 3x
4(3)=3(4)
12 = 12
checks

x2 + y2 = 25
(-
4)2 + (-3)2 = 25
1
6 + 9 = 25
25 = 25
checks
4y = 3x
4(-3)=3(-4)
-12 = -12
checks
 

 

Example 4:

Solve this system of equations algebraically:
                 x2 + y2 = 26 
(quadratic equation)
                 x - y = 6     
    (linear equation)

Substitute from the linear equation into the quadratic equation and solve.

Find the y-values by substituting each value of x into the linear  
equation.


Answer:

{(5, -1), (1, -5)}

There are 2 "possible" solutions for the system:  (4,3) and (-4,-3)
Check each in both equations.

x2 + y2 = 26
5
2 + (-1)2 = 26
25
+ 1 = 26
26 = 26
checks
x - y = 6
5 - (-1) = 6
6 = 6
checks

x2 + y2 = 26
1
2 + (-5)2 = 26
1
+ 25 = 26
26 = 26
checks
x - y = 6
1 - (-5) = 6
6 = 6
checks