Quadratic equations
Clement Radcliffe, Contributor

In this week's lesson we will continue to review quadratic equations, using both the formula and completion of squares method. Let us begin with the solution to the homework given last week.

• Solve x² - 4x - 8 = 0

x = (- b ± square root of b² - 4ac)/2a

From the equation, a = 1, b = - 4 and c = - 8. Substituting

... x = (- - 4 ± (-4)2 - 4 x 1 x -8)/2 x 1

... x = (4 ± square root of 16 + 32)/2

... x = (4 ± square root of 48)/2

... x = (4 ± 6.93)/2

... x = (4 +6.93)/2 = 10.93/2 = 5.47

OR x = (4 - 6.93)/2 = -2.93/2 = - 1.47

Answer: x = 5.47 and - 1.47

• Solve 2x² - 3x - 4 = 2 - 4x

First simplify the equation:

2x² - 3x + 4x - 4 - 2 = 0

... 2x² + x - 6 = 0 Factorising

(2x - 3)(x + 2) = 0

... 2x - 3 = 0, that is x = 3/2

And x + 2 = 0, that is x = -2.

Answer: x = 3/2 and -2

NB. If you had not realised that 2x² + x - 6 could be factorised, using the formula method would still provide the correct answer and all the marks

• Solve 2x² + 5x = 9

Expressing the equation in the required format, then

2x² + 5x - 9 = 0

Since : x = (-b ± square root of b² - 4ac)/2a

From the equation, a = 2, b = 5 and c = - 9. Substituting:

... x = (-5 ± (5)² - 4 x 2 x -9)/2 x 2

... x = (-5 ± square root of 25 + 72)/4 = (-5 ± square root of 97)/4

... x = (-5 ± 9.85)/4

... x = (-5 + 9.85)/4 = 4.85/4 = 1.21

OR x = (-5- 9.85)/4 = -14.85/4 = - 3.71

Answer: x = 1.21 OR - 3.71

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

... (x + d) = ± square root of k

... x = - d ± square root of k

As was the case with the formula method, the ± (plus or minus) will enable you to find the two values of x.

• Example: Solve x² + 4x - 3 = 0, using completion of squares.

Given x² + 4x - 3 = 0, the first critical step is to transfer the constant to the right-hand side (RHS):

... x² + 4x = 3

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 + 4x, then 4/2² or 2², the square of half the coefficient of x is required to make the Left-hand side a perfect square.

Since x² + 4x = 3, then adding 4 on both sides

... x² + 4x + 4 = 3 + 4 = 7.

NB 4 is added to both sides of the equation so that the value of x remains unchanged.

... x² + 4x + 4 = 7. (Factorising the LHS)

... (x + 2)² = 7. Find the square root of both sides

... x + 2 = ± square root of 7 = ± 2.65

... x + 2 = 2.65, that is x = 0.65

OR x + 2 = -2.65, that is x = - 4.65

I am sure you won't mind us attempting the following together.

Example

Solve x² - 8x = 1

Solution If x² - 8x = 1

Completing the squares:

... x² - 8x + - 8/2² = 1 + - 8/2² NB We add the square of half the coefficient of x.

... x² - 8x + 16 = 1 + 16

... x² - 8x + 16 = 17. Factorising

... (x - 4)² = 17

... x - 4 = ± square root of 17

... x = 4 ± 4.12

... x = 8.12 OR - 0.12

Let us now try the following together.

Solve the equation 2x² - 8x - 3 = 0, using the completion of squares method.

Solution

2x² - 8x - 3 = 0

... 2x2 - 8x = 3. I do recommend that you always make the coefficient of x² be One (1).

... Dividing by 2.

... x² - 4x = 3/2

To complete the squares, we add (- 4/2) to both sides.

... x² - 4x + - 4/2² = 3/2 + - 4/2

... x² - 4x + 4 = 3/2 + 4 = 11/2

... (x -2)² = 11/2 = 5.5 Finding the square root

... x - 2 = ± square root of 5.5 = ± 2.35

... x = 2 + ± 2.35

... x = 2 + 2.35 = 4.35

OR x = 2 - 2.35 = - 0.35

Answer: x = 4.35 OR - 0.35

Set out below is your homework.

Solve the following equations, using the completion of squares method.

(1) x² - 3x + 1= 0

(2) 3y² - 28y + 9 = 0

(3) 2x² + 2x - 8 = 3x - 6

(4) (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.

We will check your answers next week. Until then, be good.

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