|
Solving
quadratic equations
Clement
Radcliffe, Contributor
 |
| Munro
College's substitute goalkeeper
Lenwell McBean (with hand on ball),
turning the ball under the cross
bar to break up a Cornwall College
goal. Action from the daCosta
Cup game between the two schools
at Jarrett Park on November 23,
2005. Cornwall won the game 7-0.
|
LESSON
16:
This
week, we will continue the review of
solution of Quadratic Equations using
the completion of squares method.
POINTS
TO NOTE
Given
the equation x²
+ bx + c = 0
*
The constant must be shifted to the
Right Hand Side.
*
The square of half the coefficient
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.
EXAMPLE
Express
the equation x²
+ 8x = 9 in the form (x + a)²
= b.
As
the square of half the coefficient
of x must be added to both sides,
then given the equation x2 + 8x =
9, the coefficient of x is 8.
*
x²
+ 8x + (8/2)2 = 9 + (8/2)²
*
x2 + 8x + 16 = 9 + 16 = 25
*
(x + 4)²
= 25.
Let
us proceed to work together the homework
given last week.
*
Solve the equation x²
+ 9x - 4 = 0 using the completion
of squares method.
Shift
the constant to the RHS
*
x²
+ 9x = 4 Completing the squares
*
x²
+ 9x + ( 9/2)²
= 4 + (9/2)²
*
(x + 9/2)²
= 97/4 = 24.25
Taking
square root of both sides
*
x + 4.5 = ± 4.92
*
x + 4.5 = 4.92 * x = 0.42
or
x + 4.5 = -4.92 * x = -9.42
Answer
x = 0.42 or -9.42
*
Solve x²
= 7x + 5 using completion of squares.
Converting
the equation to the usual form.
*
x²
- 7x = 5
As
the coefficient of x is -7, then we
add
(-7/2)²
to both sides.
*
x²
-7x + (-7/2)²
= 5 + (- 7/2)²
= 5 + 49/4
*
(x -7/2)²
= 69/4 = 17.75 Taking square root
*
x - 3.5 = ± 4.15
*
x = 3.5 + 4.15 = 7.65
OR
x = 3.5 - 4.15 = - 0.65
Answer:
x = 7.65 or - 0.65
*
Solve the equation x²
+ 8x - 9 = 0 using completion of squares.
From
the example above, given the equation
x²
+ 8x - 9 = 0: then
x²
+ 8x = 9
*
x²
+ 8x +16 = 9 +16
*
(x + 4)²
= 25 (find the square root of both
sides)
*
(x + 4) = ±5
*
x + 4 = 5 x = 1
or
x + 4 = -5 x = - 9
Answer:
x = 1 or - 9
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.
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
factorisation method -
(x
+ B)(x - 2) = 0
*
x = 2 and - 6. Substituting in equation
(2),
*
y = 3 and 11. Answers: x = 2, y =
3 and x = - 6, 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 working.
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:
*
x²
+ 9y²
= 37
x
- 2y = -3
*
x + y = 5
xy
= 6
Have
a good week.
* Clement Radcliffe is principal
of Glenmuir High School in Clarendon.
|
