Solving quadratic equations
Clement Radcliffe, Contributor

Food and nutrition teacher, Paulette Kerr, guides students during a theory session at the Frome Technical High School in Westmoreland, recently. - Claudine Housen Photo

YOU WERE asked last week to attempt some practice examples. The following are the solutions to some of these.

Solve the simultaneous equations:

2x - 2y = 1 ... (1)
7x + 2y = 17 ... (2) Add equations (1) + (2)
ie. 9x = 18
ie. x = 2 Substitute x = 2 into equation (1)
ie. 4 - 2y = 1
ie. - 2y = 1 - 4 = -3
ie. y = 3/2

Answer: x = 2, y = 3/2

Solve simultaneously:
x + y = 7 ... (1)
2x + y = 10 .. (2)

I am sure that you can show that by subtracting equation (2) from equation (1), x = 3 and y = 4.

Solve the simultaneous equations:

- x + 3y = - 7 ... (1)
3x - 2y = 7 ... (2)

Using the elimination method, you need to convert the coefficients of one of the variables of x to 3. (- 3 is acceptable).

ie. - x + 3y = - 7 ... (1)
3x - 2y = 7 ... (2)

Multiply equation (1) by 3.
- 3x + 9y = - 21. . . . (3)

Adding equations (2) and (3)

ie. 7y = - 14
ie. y = - 2 Substituting into (2)
ie. 3x + 4 = 7
ie. 3x = 3
ie. x = 1

Answer: x = 1 and y = - 2.

If you have encountered any difficulty, you are advised to refer to last week's lesson.

Now to a new topic - SOLUTION OF QUADRATIC EQUATION (Factorisation 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 factorised.

* 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 look at some examples.

1. Solve 3x² - 7x - 6 = 0

Using factorisation:

3x² - 7x - 6 = 0
ie. (3x + 2) (x - 3) =0
ie. 3x + 2 = 0, that is, 3x = 2
ie. x = - 2/3

When x - 3 = 0,
ie. x = 3

Answer: x = -2/3 and 3

2. Solve the equation:

1 - 9x2 = 0

Factorising using difference of two squares:

1 - 9x² = (1 - 3x)(1 + 3x)

ie. (1 - 3x)(1 + 3x) = 0
ie. 1 - 3x = 0 or x = 1/3
1 + 3x = 0 ie.3x = -1
ie . x = - 1/3

Answer: x = 1/3 or - 1/3.

Alternatively

1 - 9x² = 0
ie. 9x² = 1
x² = 1/9
ie. x = ± 1/3.

3. Solve the equation: June 1990, no. 3a)

3(x +2)² = 7(x + 2)
ie. 3(x² + 4x + 4) = 7x + 14.
ie. 3x² + 12x + 12 = 7x + 14.
ie. 3x² + 12x -7x + 12 - 14 = 0.
ie. 3x² + 5x - 2 = 0. Factorizing:
ie. (3x - 1)(x + 2) = 0
ie . 3x -1 = 0, that is, x = 1/3.
OR x + 2 = 0, that is, x = -2.

Answers are: x = 1/3 and -2.

Now that you are comfortable with solving simultaneous linear equations and some quadratic equations, you can now attempt the following for homework.

• x² - 9x + 14 = 0
• 2x² - x - 15 = 0
• 2x² - x - 3 = 0
• x² + x = 6
• Solve the equation: y = 2x² - 3x - 2 when y = 0

It is important that you practise questions from CXC past papers. If you do encounter difficulties, please send your queries to me in care of The Gleaner Company.