Quadratic factors and formula
Clement Radcliffe,Contributor

If you have been following the materials presented in the last three lessons, you should realise by now that the following methods are commonly used to solve quadratic equations. These are:

• Quadratic factors
• Quadratic formula

Learning each method is important. It is also critical that you know when to use the different methods. Let us review the materials presented previously, with this in mind.

• Only some quadratic equations can be solved by the factorisation method.
• All quadratic equations wih real roots (equations with real numbers as their solutions) can be solved using the formula method
• Given the quadratic equation, you should first use the factorisation method, unless otherwise directed.
• If a specific method is requested, you must obey the instructions or you will be penalised.
• Be sure to use the correct formula and be careful in processing the negative signs in using the formula method.
• If you are asked to solve a quadratic equation correct to two decimal places, then you should use the formula method.

Please continue to practise solving quadratic equations by attempting the following:

1. Solve the equation: a² - 8a + 16 = 0

2. Solve the quadratic equation: 3x² - 5x - 4 = 0, giving your answer correct to two decimal places

3. Solve the quadratic equation: 2x² - 4x + 1 = 5x + 4

Let us now turn our attention to the homework fromt he previous lesson.

Example 1

Solve: x² + 9y² = 37

x - 2y = -3

SOLUTION

x² + 9y² = 37 ....... (1)
x - 2y = -3 ............(2)

The substitution method is used and from equation (2)

x = 2y - 3 .............(3)

Substituting Equation (3) into Equation (1)

... (2y - 3)² + 9y² = 37
... 4y² - 12y + 9 + 9y² = 37
... 13y² - 12y + 9 - 37 = 0
... 13y² - 12y - 28 = 0
... 13y² - 12y - 28 = 0
... (13y + 14)(y - 2) = 0
... 13y + 14 = 0

y = -14/13

And y - 2 = 0 .......... y = 2

Substituting into Equation (2)
Since y = -(14/13)

... x = 2 x -(14/13) - 3

= -(28/13) - 3 = -(28 - 39)/13

Ans when y=2

x = 2 x 2 - 3 = 1

Answer x = -(67/13) , y = -(14/13) and x=1, y = 2

(2) Solve the Simultaneous equation

x + y = 5
xy = 6

Solution

x + y = 5 ......... (1)
xy = 6 ............. (2)

From equation (1) x = 5 - y (3)

Substitute into equation (3) with equation (2)

... (5 - y) x y = 6

... 5y - y² = 6
... y² - 5y + 6 = 0
(y - 3) (y - 2) = 0

y - 2 = 0, y = 2

When y = 3 substitute in equation (3)

... x = 5 - 3 = 2

When y = 2 ....... x = 5 - 2 = 3

... Answer: x = 3, y = 2 and x = 2, y = 3

We will now complete ALGEBRA by reviewing aspects of GRAPHS

GRAPHS

Please be reminded that you are required to be ale to draw straight line and quadratic graphs.
In doing so, it is important that you pay attention to the following:

• You need to complete accurately an appropriate table of X and Y values
• The X and Y axes must be CLEARLY LABELLED
• The scale used must be appropriate to the problem. If one is given, it must be accurately used.
• A ruler must be used to draw the straight line while free hand must be used to draw the curve
• The use of a suitable pencil (HB) is required

APPLICATIONS

Graphs may be used to solve:

• Quadratic equations
• Simultaneous equations

In both cases, the solution is represented by the X and Y coordinates of the points of intersection of the line and the curve.

EXAMPLE

Plot the equations y = 3x² - 2x - 1 and y = x + 5

Hence: (a) Solve the equation 3x² - 2x - 1 = 0.

(b) Solve both equations simultaneously

Completing the tables:

y = 3x² - 2x - 1

y = x + 5

(a) The solution of 3x² - 2x - 1 = 0 is the X coordinates of the points of intersection of the curve and the X axis.

We will continue the review of graphs next week.

Clement Radcliffe is principal of Glenmuir High School. Send questions and comments to kerry-ann.hepburn@gleanerjm.com