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Reviewing
quadratic equations
Clement Radcliffe, Contributor
 | Sangster's
Book Stores promotions representative, Yvonne Christie (left), looks through a
Collins Dictionary with Lacovia Comprehensive High School students (from left)
Kerene Kirlew (head girl), Everie Salmom (head boy), and Rhonette Robinson (students'
council president) following the handing over of more than $100,000 worth of books
to the school's library under the Sangster's Book Store Outreach Programme. -
Claudine Housen/Staff Photographer |
In
this week's lesson, we will continue to review quadratic equations, using both
the formula and completion of squares methods. Let's begin with the solution to
the homework given last week. Solve
2x² -
6x - 1 = 0
 Quadratic
equations may also be solved using the method of Completion of Squares. Solve
2x²
- 3x - 4 = 2 - 4x First
simplify the equation: ie.
2x² -
3 + 4x - 4 - 2 = 0 ie.
2x² +
x - 6 = 0 Factorizing (2x
- 3)(x + 2) = 0 ie.
2x - 3 = 0, that is x = 3/2 And
x + 2 = 0, that is x = -2 Answer:
x = 3/2 and -2 Quadratic
equations may also be solved using the method of Completion of Squares. COMPLETION
OF SQUARES Given
the equation x²
+ bx + c = 0, the aim is to convert the equation to the form (x + d)²
= k, where d and k are constants. You are, therefore, required to
Convert the left hand side to a Perfect Square of form (x + d)².
Given the form (x + d)²
= k, then x is found by determining the square root of both sides. (x
+ d)²
= k ie.
(x + d) = ±  k ie.
x = -d±k As
was the case previously, the ± (plus or minus sign) will enable you to
find the two values of x. Example: Solve
x² + 6x
- 5 = 0, using completion of sqaures. Given
x² + 6x
- 5 = 0, the first critical step is to transfer the constant to the Right Hand
Side (RHS): ie.
x² + 6x
= 5 This
is followed by making the Left Hand Side a perfect square. This is based on the
following equation: (x
+ a)² = x²
+ 2ax + a² Given
x² + 2ax,
then a²,
the square of half the coefficient of x must be added to complete the square. Given
x² + 6x
= 5, then [6/2]²
or 3²,
the square of half the coefficient of x is required to make the Left Hand Side
a perfect square. Since
x² + 6x
= 5, then adding 9 on both sides ie.
x² + 6x
+ 9 = 5 + 9 = 14 N.B.
9 is added to both sides of the equation so that the value of x remains unchanged. | ie.
x² + 6x
+ 9 = 14 | (Factorixing
the LHS) | | ie
(x + 3)² = 14 | Find
the square root of both sides | | ie.
x + 3 = ± 14
= ± 3.74 | | ie.
x + 3 = 3.74, that is x = 0.74 | | OR
x + 3 = -3.74, that is x = -6.74 | | | Answer:
x = 0.74 or -6.94 |
I
am sure you won't mind us practising the following together. Examples (1)
Express x²
- 8x = 1 in the form (x+a)²
= b As
[8/2]²
= 16, then 16 must be added to both sides to make the LHS a perfect square. | ie.
x² - 8x + 16 = 1 + 16 | Factorizing
the LHS | | ie.
(x-4)² = 17 | |
You
may expand to check your answers. (2)
Find the values to be added to the LHS to make it a perfect square. (a)
x² - 12x
= 1
(b) y²
+ 9x = 4 Your
answers should be 36 and 81/4 respectively. If this is not so, then please review
the method. Now
let us solve the above examples together. (a)
Solve x²
- 8x = 1
If x²
- 8x = 1 Completing
the squares: ie.
x² - 8x
+ 16 = 1 + 16
ie.
(x - 4)²
= 17
ie. x - 4 = ±17
ie.
x = 4 ± 4.12
ie. x = 8.12 and -.12 (b)
Solve x²
- 12x = 1
Since x²
- 12x = 1
Completing
the squares: ie.
x² - 12x
+ 36 = 1 + 36
ie. (x-6)²
= 37
ie. x = ±37
ie.
x = 6± 6.08
ie. x = 12.08 and -.08 Set
out below is your homework. Solve
the following equations, using the completion of squares method. (1)
x² + 9x
- 4 = 0
(2) x²
+ 8x - 9 = 0
(3) x²
= 7x + 5 Clement
Radcliffe is the principal of Glenmuir High School in May Pen. | 
