Quadratic equations
Clement Radcliffe, Contributor

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.
• This constant must be added to 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 below. But first, let us do an example together.

Example

Express the equation x² - 3x + 1 =0 in the form (x + a)² = b. Hence solve the equation.

(i) x² - 3x + 1 =0
... x² - 3x = -1

As the square of half the coefficient of x must be added to both sides, then given the equation x² - 3x = -1, the coefficient of x is -3.
... x² - 3x + (-3/2)² = -1 + (-3/2)²
... x² - 3x + 9/4 = -1 + 9/4 = 5/4 by factorizing the Left Hand Side
... (x - 3/2)² = 5/4

(ii) In order to solve the equation, we first find the square root of both sides:

... (x - 3/2) = + or - square root of 1.25 = + or - 1.12
... x - 3/2 = 1.12
... x = 1.5 - 1.12 = 2.62

OR x - 3/2 = -1.12
... x = 1.5 - 1.12 = 0.38

Answer: x = 2.62 OR 0.38

Solution to Homework

Solve 2x² + 2x - 8 = 3x - 6
... 2x² + 2x - 8 = 3x - 6
... 2x² + 2x - 3x = 8 - 6
... 2x² - x - 8 = 2 Dividing by 2
... x² - x/2 = 1

As -1/2 is the coefficient of x, then (-1/4)² is added to both sides.

... x² - x/2 + (-1/4)² = 1 + (-1/4)²
... x² - x/2 + 1 + 1/16 = 17/16

Factorizing the Left Hand Side
... (x - 1/4)² = 17/16 = 1.06

Taking the square root of both sides

... (x - 1/4) = + or - square root of 1.06 = + or - 1.03
... x - 0.25 = 1.03
... x = 1.28
OR x - 0.25 = 1.03
... x = -0.78

Answer x = 1.28 OR -0.78

Let us now consider another example.

Example

(i) By simplifying, show that (2x - 3)(2x + 3) - (x - 4)² = (3x² + 8x - 25)

(ii) Write (3x² - 8x - 25) in the form a(x + h)² + k, where a, h and k are real numbers.

(iii) Solution

(i) Given (2x - 3)(2x + 3) - (x - 4)² Clearing the brackets
... (4x² - 6x + 6x - 9) - (x² - 8x + 16)
... 4x² - 9 - x² + 8x - 16 = 3x² + 8x - 25

(ii) Given 3x² + 8x - 25

= 3 ( x² + 8/3 - 25/3) To complete the square, you add (4/3)²
= 3 (x² + (8/3)x + (4/3)² - (4/3)² - 25/3)
= 3 (x² + (8/3)x + 16/9 - 16/9 - 25/3) Simplifying
= 3 ( (x + 4/3)² - (16/9 + 25/3)
= 3 ( (x + 4/3)² - (16/9 + 75/9) = 3 ( (x + 4/3)² - 91/9)
= 3 (x + 4/3)² - 91/3
... 3x² - 8x - 25 = = 3 (x + 4/3)² - 91/3

We will continue the review of ALGEBRA by returning to the solution of Simulatenous Equations. Today, I will deal specifically with those cases in which one equation is linear and one quadratic.

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

Using the factorization method:
(x + 6)(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.

Example

Determine two numbers whose sum is 9 and whose product is 20, by solving a quadratic equation.

Let the munber be x and y.
... x + y = 9 ....(1)
x X y = 20 ......(2)

From Equation (1),
x = 9 - y ...... (3)

Substituting Equation (3) in Equation (2),

... (9 - y) x y = 20
... 9y - y² = 20
... y² - 9y + 20 = 0 Factorizing
... (y - 5)(y - 4) = 0
... y - 5 =0
... y = 5

OR y - 4 = 0
... y = 4

Substituting into Equation (1)

When y = 5
... 5 + x = 9
.... x = 4

When y = 4
... 4 + x = 9
... x = 5

Answer y=5 and x=4
OR y = 4and x = 5

Please attempt to solve the following on your own:

x² + 9y² = 37
x - 2y = -3

y - x =1
y = x² - 3x +4

Enjoy the rest of the week.

Solve the following simulatneous equations:

y = x² - 3x + 4
y - x =1

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Clement Radcliffe is the principal of Glenmuir High School in May Pen.