Reviewing quadratic equations - II
Clement Radcliffe, Contributor

Frome Technical High School students play a game of chess during break recently. - Claudine Housen/Staff Photographer

This week, we will continue the review of the solution of Quadratic Equations using the completion of 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 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.

Given the equation ax² + bx + c = 0

• The initial step is ax² + bx = -c
• Then divide by a to make the coefficient of x one, that is x² + b/a x = -c/a

The method continues as above.
The above is illustrated by the solution to the Homework as follows:

Example

Express the equation x² + 8x = 9 in the form (x + a)² = b
Hence solve the equation

(i) x² + 8x = 9

As the square of half the coefficient of x must be added to both sides, then given the equation

ie. x² + 8x+ (8/2)² = 9 + (8/2)²
ie. x² + 8x + 16 = 9 + 16 = 25
ie. (x + 4)² = 25

(ii) In order to solve the equation, we first find the square root of both sides:
ie. (x + 4) = ±5
ie. x + 4 = 5 ie. x = 1
OR x + 4 = -5 ie. x = -9

Answer: x = 1 or -9

• Solve x² = 7x + 5 using the completion of squares method

Converting the equation to the usual form
ie. x² - 7x = 5

As the coefficient of x is -7, then we add (-7/2)² to both sides.

ie. x² - 7x + (-7/2)² = 5 + (-7/2)² = 5 + 49/4

OR x = 3.5 - 4.15 = -0.65

Answer: x = 7.65 or -0.65

• Solve the equation x² + 9x - 4 = 0 using the completion of squares method

Shift the constant to the RHS

ie. x + 4.5 = ±4.92
ie. x + 4.5 = 4.92 ie. x = 0.42
or x + 4.5 = -4.92 ie. x = -9.42

Answer x = 0.42 or -9.42

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.

SIMILTANEOUS 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 in equation (1),
5 - x = x² + 3x - 7
x² + 4x - 12 = 0

Using factorization method
(x + 6)(x - 2) = 0
ie. x = 2 and -6 Substituting in equation (2),
ie. 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 - 3y² = 20 ......... (1)
  2x + y = 6 ................(2)
  From Equation (2),
  y - 6 - 2x .................(3)

Substituting Equation (3) in Equation (1),

Substituting into Equation (2)

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 the principal of Glenmuir High School in May Pen.