Lesson Defining Quadratic Equations Given the Solutions

Source code of 'Defining Quadratic Equations Given the Solutions'

This Lesson (Defining Quadratic Equations Given the Solutions) was created by by oberobic(2304)

: View Source, Show
About oberobic: MBA/Ph.D. University Administrator

Problem: We are given the solutions (zeroes) of a quadratic equation and asked to define the equation that has those solutions. In this case, we are told the solutions (zeroes) are -5 and 2. 
.
Solution: Given the zeroes, the quadratic equation can defined as follows:
.
The concept of the zeroes is that these are the values of 'x' that will result in a product of 0.
.
a(x+5)(x-2) = 0
.
If (x+5)=0, then x=-5.
If (x-2)=0, then x=2.
These are the values we were given.
.
However, we cannot say what the value of 'a' is unless we have some additional information.
'a' is simply a constant that would affect the graph's position.
.
Expand using FOIL.
.
(x+5)(x-2) = x^2 +3x -10
.
Now we can graph the equation, but we need to know the value of 'a' to determine which if the infinitely many graphs is the answer.
.
In the following graph:
The red line assumes 'a' = 1.
The green line assumes 'a' = -1.
The blue line assumes 'a' = 1/2.
The purple line assumes 'a' = -1/4.
.
{{{ graph(500,500,-10,10,-20,20,x^2+3x-10,-1*(x^2+3x-10),1/2*(x^2+3x-10),-1/4*(x^2+3x-10) ) }}}
.
To decide which of these is the solution, you would need to be given the y-intercept. Then you could experiment with values of 'a' to cross at the intercept.
.
Answer: a(x+5)(x-2) = 0. Additional information is needed to solve for 'a'.