|
Quadratic
equations using graphs
Clement Radcliffe, Contributor
 |
| Students
visit a booth at the Ministry of Education and Youth's inner-city Schools Improvement
Programme and Beacons of Progress and Achievement capacity-building workshop at
the Stella Maris church hall on Tuesday, January 30. - Rudolph Brown/Chief Photographer |
This week
we will complete the review of algebra by considering aspects of graphs. Specifically,
I will elaborate on the solution of quadratic equations using a graph. Reminders
- A
quadratic equation is represented graphically by a curve.
- A
curve should be drawn by freehand sketch.
- The
x axis has the equation y = 0 and the y axis has equation X = 0.
- Given
the curve y = f(x) and the line y =g(x), then the points of intersection of both
are represented by:
y
= f(x) = g(x) ie. f(x) = g(x) If
f(x) =x²
- 3x + 2 and g (x) = 2x -1 Then,
at the point of intersection of the curve and the line f(x) = g(x). ie.
x² - 3x
+ 2 = 2x -1
ie.
x² - 3x
- 2x + 2 + 1 = 0
ie.
x² - 5x
+ 3 = 0 The
x coordinates of the points of intersection is therefore the solution of the equation
x² - 5x
+ 3 = 0 Example
Using
an appropriate scale, please plot the curve y = 3x²
- 2x - 1. Hence, solve the equations: a)
3x² -
2x - 1 = 0
b)
3x² -
1 = 2 -2x
c)
3x² -
3 = 0 or x2 - 1 = 0 
a)
Given the curve y = 3x²
- 2x - 1, the solution of the equation 3x²
- 2x - 1 = 0 is the x values of the points of intersection of the curve y = 3x²
- 2x -1 and the line y = 0 or the x axis. ie.
The solution is x = 1, -.33 b)
Given the curve y = 3x²
- 2x -1 by plotting the line y = 2 - 2x, then the points of intersection of the
curve and the line will represent the solution of the equation 3x²
- 2x - 1 = 2 -2x. From
the graph the solution is x = -1, 1. c)
Given the equation 3x²
- 3 =0, if the curve y = 3x²-2x-1
must be used, then 3x²
- 3 = 0 is reorganised as follows: 3x²
-2x + 2x-2 - 1 = 0 ie.
3x² -2x
-1= 2 - 2x ie.
The solution of the equation 3x²
- 3 = 0 is the x coordinates of the points of intersection of the curve y = 3x²-2x-1
and the line y = 2-2x. ie.
As in b) the solution is x = -1,1. Let's
attempt another example. Given
the curve y = 2x²-x-3,
solve the equation 2x²
-2x-5 = 0. By
reorganising the equation 2x(2) -2x-5 = 0, it follows that: 2x²-x-x-3-2
= 0 ie.
2x²-x-3
= x+2 Then,
the solution of 2x²
-2x-5 = 0 is the x coordinates of the points of intersection of the curve y =
2x² -x-3
and the line y = +2. Please
continue to solve similar examples from past papers. This section two question
must be done if it appears on your exam paper in June. Clement
Radcliffe is the principal of Glenmuir High School in May Pen. | 
