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Quadratic
equations
By Clement Radcliffe, Contributor
IN
OUR previous lesson, I reviewed the
solution of quadratic equations, using
the graphical method. The following
points bear repeating:
Given
the graph of the curve y = 3x²-2x-1;
*
The solution of the equation 3x²-2x-1
= 0 is the x coordinates of the points
of intersection of the curve and the
x axis.
*
The solution of the equation 3x²-2x-1
= 2 or 3x²-2x-3 = 0 is the x
coordinates of the points of intersection
of the curves and the line y = 2.
*
The solution of the equation 3x²-2x-1
= 4-x or 3x²-x-5 = 0 is the x
coordinates of the points of intersection
of the curve and the line y = 4-x.
The
above may be illustrated by the following
example:
Solve
graphically the simultaneous equations:
y
= 3x²-2x-1 and y = x+5
SOLUTION
Completing
the tables:
| y
= 3x²-2x-1 |
y
= x+5 |
| x
-3 -3 -1 0 1 2 3 |
x
0 1 2 |
| y
32 15 4 -1 0 7 20 |
y
5 6 7 |
The
points of intersection are: (-1, 4)
and (2, 7).
The
solutions are: x = -1, y = 4 and x
= 2, y = 7
Please
note that at the points of intersection
of the curve y = 3x²-2x-1 and
the line
y
= x+5, then 3x²-2x-1 = x+5.
...3x²-3x-6
= 0 or x²-x-2 = 0
The
solution of the simultaneous equations
is also the solution of the above
equations.
I
am sure that you realise that the
above graph can also be used to solve
the equation 3x2-x-3 = 0. If you have
a doubt, then please note the following:
3x²-x-3 = 3x²-2x+x-1-2 =
0
=
3x²-2x-1 = 2-x.
The
solution of the equation 3x²-x-3
= 0 is the points of intersection
of the curve y = 3x²-2x-1 and
the line y = 2-x.
Now
let us continue our review of another
aspect of Coordinate Geometry with
the solution to the homework given
last week.
HOMEWORK
Find
the gradient, midpoint and length
of the line joining the points.
(a)
A (-3,-4), B (3,7) (b) C (8,5), D
(-4,2)
SOLUTION
The
gradient is found using the formula:
m = y2 - y1 ÷ x2 - x1
The
gradient of A (-3,-4), B (3,7) is
found by substituting.
...m
= 7- -4 ÷ 3- -3 = 7 + 4 ÷
3 + 3 = 11 ÷ 6
Similarly,
the gradient of CD is m = 2 - 5 ÷
-4 - 8 = -3 ÷ -12 = 1/4
The
mid-point is found using the formula:
Given
the points A and B above,
| M
= 3-3, 7-4 = |
0,
3 |
|
2
|
2 |
2
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For
C and D above:
| M
= 8-4, 5+7 = |
2,
7 |
|
2
|
2 |
2
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The
formula for the length of the line
AB is =
...For
the line AB: =
For
the line CD, the length is =
I
do urge you to continue to attempt
to solve similar problems. Next week,
we will pursue aspects of the equations
of a straight line.
Clement
Radcliffee is Principal of Glenmuir
High School in Clarendon. Send your
questions and comments to the CXC
Study Guide, the Gleaner Company Ltd.,
7 North Street, Kingston; or email
us at jcampbell@gleanerjm.com.
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