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Quadratic
Equations (Cont'd)
ClementRadcliffe, Contributor
This week, we will continue the review
of the solution of Quadratic equations
using the completion squares method.
Points
to note
Given the equation x2+bx+c
= 0, you are asked to note the following
with respect to the completion of
squares method.
- The
constant must be shifted to the
Right Hand Side.
- The
square of half the coefficent of
x must be added to the Left Hand
Side to make it a perfect square.
- The
constant must be added
to both sides of the equation.
- The
value of x is determined by finding
the square root of both sides of
the equation.
- The
value of x is determined by finding
the square root of both sides of
the equation.
Given
the equation ax2+bx+c =
0
The initial step is ax2+bx
= -c
Then
divide by a to make the coefficent
of x one, that is x2+(b/a)x
= -c/a
The
method continues as above.
The
above method is illustrated by the
solution to the Homework below. But
first, let us do an example together.
Example
Express
the equation x2 -3x +1
= 0 in the form(x+a)2 =
b. Hence solve the equation.
(i) x2 -3x +1 = 0
... x2 -3x = -1
As the square of half the coefficent
of x must be added to both sides,
then given the equation
x2 -3x = -1, the coefficent
of x is -3.
... x2 -3x + (-3/2)2
= 1 + (-3/2)2
... x2 -3x + 9/4 = -1 +
9/4 = 5/4
... (x - 3/2)2 = 5/4
(ii) In order to solve the equation,
we must first find the squre root
of both sides:
... (x
- 3/2)
= (+) or (-) ?1.25 = (+) or (-) 1.12
... x
- 3/2
= 1.12 ... x = 1.5 + 1.12 = 2.62
OR x
- 3/2
= 1.12 ... x = 1.5 - 1.12 = 0.38
Answer:
x = 2.62 OR 0.38
Solution
to Homework
Solve 2x2 + 2x - 8 = 3x
- 6 using the completion of squares
method
Since 2x2 + 2x - 8 = 3x
- 6
... 2x2 - x = 2 (Dividing
by 2)
... x2 - x/2 = 1
As -1/2 is the coefficent of x, then
(-1/4)2 is added to both
sides.
... x2 - x/2 + (-1/4)2
= 1 + (-1/4)2
... x2 - x/2 + 1/16 = 17/16
Factorizing the Left Hand Side
... (x - 1/4)2 = 17/16
= 1.06
Taking the square root of both sides
... (x - 1/4) = (+) OR ( - ) ?1.06
= (+) OR ( - ) 1.03
... x - 0.25 = 1.03 ...x = 1.28
OR x - 0.25 = -1.03 ...x = -0.78
Answer: x
= 1.28 OR -0.78
Let us now continue the review of
ALGEBRA
by returning to the solution of Simultaneous
Equations. Today, I will deal specifically
with those cases in which one equation
is linear and one quadratic.
Simultaneous Equations - One
linear and one quadratic
The substitution method is used.
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 = 5 - x in equation
(1)
5 - x = x2 + 3x - 7
... x2 + 3x + x - 7 - 5
= 0
... x2 + 4x - 12 = 0
Using the factorization method:
(x + 6)(x - 2) = 0
... x = 2 and -6. Substituting
in equation (2)
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
na error has been made. It is therefore
recommended that you check your working.
Let us do another example together.
Example
Determine
two numbers whose sum is 9 and whose
product is 20, by solving a quadratic
equation.
Let the numbers be x
and y
... x + y = 9 ......9 (1)
x*y = 20 .............(2)
From equation (1)
x = 9 - y
Substituting Equation (3) in Equation
(2)
...(9 - y) * y = 20
...9y - y2, =20
... y2 - 9y + 20 = 0 (Factorizing)
... (y - 5)(y - 4) = 0
... y - 5 = 0 ... y = 5
OR y - 4 = 0 ... y = 4
Substituting into equation (1)
When y = 5 ... 5 + x = 9 ... x = 4
When y = 4 ...4 + x = 9 ... x = 5
Answer: y = 5 and x = 4
OR y = 4 and x = 5
Please attempt to solve the following
on your own:
(a)
x2 + 9y2 = 37
x - 27 = -3
(b) x + y = 5
xy = 6
Enjoy the rest of the week.
Clement Radcliffe is the principal
of Glenmuir High School in May Pen.
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