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Solution
of quadratic equations
Clement Radcliffe,Contributor
We
will complete our review of simultaneous
equations this week by looking at
the solutions to some of the practice
examples that were given for homework.
Solve
simultaneously:
3x
+ 2y = 1 . . .(1)
4x
- y = 16 . . . (2)
Multiply
equation (2) by 2
8x
- 2y = 32. (3)
Add
equations (1) and (3)
11x
= 33
x
= 33⁄11
= 3.
Substituting
x = 3 into equation (1):
3
x 3 + 2y = 1
9
+ 2y = 1
2y
= 1 - 9 = - 8
y
= - 4
Answer:
x = 3, y = -4.
Solve
simultaneously:
2x
= 11 + 3y . . . (1)
x
+ 2y + 12 = 0 . . . (2)
Using
the substitution method: From equation
(1), x = ( 11 + 3 )/2
Substituting
into equation (2):
(
11 + 3y )/2 + 2y +12 = 0. Multiply
all terms by 2 to clear the 2 denominator:
2
x ( 11 + 3y )/2 + 2 x 2y + 2 x 12
= 2 x 0.
11
+ 3y + 4y + 24 = 0
7y
= - 24 -11 = - 35
y
= -35⁄7
= - 5
Substituting
into equation (1):
x
= ( 11 + 3y )/2 = ( 11 + 3 x - 5 )/2
= ( 11 - 15 )/2
x
= -4⁄2
= -2
Answer:
x = -2, y = -5.
Solve
simultaneously:
2x
+ 3y = 3. . . (1)
5x
- 2y = 17 . .(2)
Multiply
equation (1) by 2 and equation (2)
by 3.
4x
+ 6y = 6 . . . (3)
15x
- 6y = 51....(4)
(You
could have multiplied equation (1)
by 5 and equation (2) by 2 with similar
result, please attempt this approach.
)
Adding
equations (3) and (4)
19x
= 57
x
= 57⁄19
= 3
Substituting
into equation (3)
=
4 x 3 + 6y = 6
12
+ 6y = 6
6y
= 6 - 12
6y
= - 6
y
= - 1
Answer
x = 3 , y = - 1
Now
to a new topic: Solution of quadratic
equations
The
following are the methods which are
commonly used at this level.
- Factorisation
-
Graphs
- Formula
method
We
will now begin with the factorization
method.
POINTS
TO NOTE
- Quadratic
equations are expressed in the form
ax2 + bx + c = 0, where a, b and
c are constants.
- The
factorisation method is used if,
and only if, the expression ax2
+ bx + c can be factorized.
- Given
the equation x2 + 7x + 10 = 0, then
by factorising the left hand-side
you get
(x
+ 2 )( x + 5 ) = 0.
If
(x + 2 )( x + 5 ) = 0
then
(x + 2) = 0, that is x = - 2
OR
x + 5 = 0, that is x = -5.
Solutions
are x = -2 and x = -5.
- Be
reminded that the solutions of the
equation are the values which satisfy
the equation.
- These
can be checked by substitution as
follows:
If
x2 + 7x + 10 = 0, then
if x = -2, 4 -14 + 10 = 0.
Similarly,
where x = -5, then 25 -35 + 10 =
0. The equation is satisfied by
both solutions.
We
shall now look at some other examples.
1.
Solve 3x2 - 7x -6 = 0
Using
factorisation:
3x2
- 7x -6 = 0
(3x
+ 2) (x - 3) =0
3x
+ 2 = 0, that is, 3x - 2
x
= - 2/3
When
x - 3 = 0,
x
= 3
Answer:
x = - ? and 3
2.
Solve the equation:
1
- 9x2 = 0
Factorising
using difference of two squares:
1
- 9x2 = (1 - 3x)(1 + 3x)
(1
- 3x)(1 + 3x) = 0
1
- 3x = 0 or x = 1/3
1
+ 3x = 0 3x = -1
x
= - 1/3.
Answer:
x = 1/3 or - 1/3.
Alternatively
1
- 9x2 = 0
9x2
= 1
x2
= 1
9
x
= ± 1/3.
3.
Solve the equation: 3(x +2)2
= 7(x + 2)
3(x
+2)2 = 7(x + 2) Clearing
the brackets:
3(x2
+ 4x + 4) = 7x + 14.
3x2
+ 12x + 12 = 7x + 14.
3x2
+ 12x -7x + 12 - 14 = 0.
3x2
+ 5x - 2 = 0. Factorizing:
(3x
- 1)(x + 2) = 0
3x
-1 = 0, that is, x = 1/3.
OR
x + 2 = 0, that is, x = -2.
Answers
are x = ? and -2.
Note
You
may also solve the above using the
factorisation method. In this case:
3(x
+2)2 = 7(x + 2) is expressed
as 3(x +2)2 - 7(x + 2)
= 0
The
common factor is ( x + 2)
Factorising ( x + 2 ) ( 3( x + 2)
- 7) = 0
(
x + 2 ) ( 3x + 6 - 7 ) = 0
(
x + 2 ) ( 3x - 1 ) = 0
I am sure you are able to complete
the solution.
Now
that you are comfortable with solving
simultaneous linear equations and
some quadratic equations, attempt
the following for homework.
1.
x2 - 3x - 4 = 0
2.
6x2 - x - 15 = 0
3.
2x2 - x - 3 = 0
4.
x2 + x = 6
5.
Solve the simultaneous equations:
3a
- 1/2b = 4
9a
+ 2b = -2
I
urge you to find other examples in
your textbooks and past papers and
do them. After all, practice becomes
perfect.
Clement
Radcliffe is an independent contributor.
Send questions and comments to kerry-ann.hepburn@gleanerjm.com
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