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Factorisation
By Clement Radcliffe, Contributor
I
WILL begin today's lesson by giving
the solution to last week's homework.
*
Solve the quadratic equation: x²
+ 3x + 1 = 0
Having
established that the factorisation
method cannot be used, then the formula
method is selected.
From
x² + 3x + 1 = 0, then a = 1,
b = 3, and c = 1.
Substituting,
...
...x
= - 3 + 2.24 ÷ 2 = -0.38
OR x = - 3 - 2.24 ÷ 2 = -2.62
Answer is x = -0.38 and -2.62
*
Solve the quadratic equation: 2x²
- 6x -1 = 0
Using
the formula method, then a = 2, b
= -6, and c = -1.
Substituting,
...x
= 6 = 6.63 ÷ 4
...x = 6 + 6.63 ÷ 4 = 3.16
OR x = 6 - 6.63 ÷ 4= -0.16
Answer is x = 3.16 and -0.16
*
Solve the quadratic equation: 6x2
+ 11x = 10
6x²
+ 11x = 10 ...6x² + 11x - 10
= 0
As
above, a = 6, b = 11, and c = -10.
Substituting
in the usual quadratic formula,
Answer
is x = 0.67 and -2.5
I
do hope that you realise that you
could have used factorisation to solve
the quadratic equation: as follows:
6x²
+ 11x - 10 = 0
Factorising:(2x
+ 5)(3x - 2) = 0.
| ...2x
+ 5 = 0 ... x = |
-
5 |
| |
2 |
...3x
- 2 = 0 ... x =2/3
*
Solve the quadratic equation: 2x2
- 3x - 4 = 2 - 4x.
Given
2x² - 3x - 4 = 2 - 4x.
...2x²-
3x + 4x - 4 - 2 = 0
...2x²
+ x - 6 = 0
Factorising:(2x
-3)(x + 2) = 0.
...2x - 3 = 0 ...x = 3/2
...x
+2 = 0 ...x = - 2.
Answer
is x = 3/2 and -2.
1
will complete the review of ALGEBRA
by returning to the solution of simultaneous
equations. I will deal specifically
with those cases in which one equation
is linear and one quadratic.
Please
be reminded that the solution of simultaneous
equations is either the values of
x and y which satisfy both equations,
or the co-ordinates of the point(s)
of intersection of lines representing
both equations.
SIMULTANEOUS
EQUATIONS One linear and one
quadratic.
EXAMPLE:
Solve
the following equations:
y
= x² + 3x - 7... (1)
y
+ x = 5 ... (2)
The
substitution method is used as follows:
From
equation (2), y = 5 - x
Substituting
y in equation (1),
5
- x = x² + 3x - 7
x²
+ 4x - 12 = 0
Using
the factorisation method-
(x + 6)(x - 2) = 0
...x = 2 and -6. Substituting in equation
(2),
...y = 3 and 11.
Answer: x = 2 and y = 3
x
= -6 and y = 11
Kindly
note the following:
(a)
There are TWO SETS of values
because of the quadratic equation.
(b)
The basic principles of Algebra should
be well known, as they are required.
If
your solutions have large values,
for example 136, it is likely that
an error has been made. It is therefore
recommended that you check your work.
Let
us do another example together.
*
Solve the following equations:
2x²
+ y² = 33 ... (1)
x
+ y = 3 ... (2)
From
equation (2), x = 3 - y
Substituting
in equation (1),
2(3
-y)² + y² = 33
...2(9
- 6y + y²) + y²= 33
...18
- 12y + 2y² + y² = 33
...
3y² - 12y + 18 - 33 = 0
3y²
- 12y - 15 = 0
...y²
- 4y - 5 = 0
...(y
- 5)(y + 1) = 0
...y
= 5 and - 1
Substituting
into equation (2) x = 3 - y
when
y = 5, x = - 2
when
y = - 1, x = 4
Answer:
x = -2 and y = 5
x
= 4 and y = -1
Please
attempt to solve the following on
your own:
*
3x + 2y = 19
xy
= 15
*
x2 + y2 = 24
y
= 2x + 3
*
x + y = 5
xy
= 6
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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