Algebra
Clement Radcliffe, Contributor

As you continue your preparation for the 2009 Summer CSEC examinations, I do hope that your efforts will not be affected by the imminent Christmas holidays.

We will continue witht he review of ALGEBRA, by solving together the following Quadratic Equations.

Solve the following:

• x² - 3x - 4 =0 Factorize
  ... x² - 3x - 4 = (x - 4)(x + 1)
  ... (x - 4)(x + 1) = 0
  ... x-4 =0, tha t is, x=4 OR x + 1 =0, that is, x= -1
  
• 6x² - x - 15 =0 Factorize
  ... (3x - 5)(2x + 3) = 0
  ... 3x-5 =0, that is, x=5/3 OR 2x + 3=0, that is, x = -3/2
• x² + x =6
  ... x² + x - 6 =0
  ... (x + 3)(x - 2) =0
  ... x = -3 and 2.
• Solve the simultaneous equations:
  3a - 1/2b = 4 ...(1)
  9a + 2b = -2 ...(2)

Multiply Equation (1) by 4
... 12a - 2b = 16 ... (3)

Add equations (2) and (3)
... 12a = 14
...a = 14
... = 21
... = 2/3

Substituting into Equation (1)
... 3 x 2/3 - 1/2b = 4
... 2 - 1/2b = 4
... -1/2b = 2
... b = -4
Answer: a = 2/3 and b = -4

Most quadratic equations cannot be solved by factorization. Alternatively, the FORMULA METHOD is used. Please be reminded that given the quadratic equation ax² + bx + c =0, where a, b and c are constants, then it can be shown that x = (-b +OR- Square root of b² - 4ac)/2a

This is the basic of the formula method as x is found by substituting teh values of a, b and c into the formula.

Examples:
Express p² - 2p - 11 =0 in the form ax² + bx + c =0 and find the values of a, b and c

By comparing this equation with the required form ax² + bx + c =0
... a = 1, b = -2 and c= -11
Please be careful not to omit th enegative sign.
Answer: a = 1, b = -2 and c= -11

• Solve the following quadratic equation expressing the value of p to two decimal places:
  p² - 2p - 11 =0

NB. The solution correct to two decimal places implies the use of the formula method.

... p= (-b +OR- Square root of b² - 4ac)/2a

From the given equation, a = 1, b = -2 and c= -11.

Substituting

...p= (-(-2) +OR- Square root of (-2)² - 4 x 1 x (-11))/2 x 1
... p= (2 +OR- Square root of - 4 + 44)/2 = (2 +OR- Square root of 48)/2
... p= (2 +OR- 6.93)/2
... p= (2 + 6.93)/2 P = (8.93)/2 = 4.47
OR p= (2 - 6.93)/2 = (4.93)/2 = -2.47

Answer is p= 4.47 OR -2.47

Let us try another example.

• Express 2x² = 3x + 1 in the form ax² + bx + c =0 and find the values of a, b and c

Hence solve te quadratic equation.

Given that 2x² = 3x + 1, then 2x² = 3x + 1 =0.

By comparing this equation with the required form ax² + bx + c =0
... a= 2, b = -3 and c= -1

Given the formula x = (-b +OR- Square root of b² - 4ac)/2a, then substituting

...x= (-(-3) +OR- Square root of (-3)² - 4 x 2 x (-1))/2 x 2
... x = (3 +OR - Square root of 9 + 8)/4
... x = (3 +OR- Square root of 17)/4 = (3 + OR- 4.12)/4
... Either x = (3 + 4.12)/4 = (7.12)/4 = 1.78
Or x = (3 - 4.12)/4 = (-1.12)/4 = -0.28
Answer is x= 1.78 OR -0.28

We will work another example together.

Solve the following equation using the quadratic formula: 2x² + 2x -8 = 3x - 6
2x² + 2x -8 = 3x - 6
2x² + 2x -8 = 3x - 6 = 0
2x² - x - 2 = 0

Having expressed the equation into the appropriate form, then a = 2, b = -1 and c = -2

Using the formula x = (-b +OR- Square root of b² - 4ac)/2a

...x= (1 +OR - Square root of 1 - 4 x 2 x -2)/4 = (1 + OR - Square root of 1 + 16)/4

... x = (1 +OR- Square root of 17)/4 = (1 + OR- 4.12)/4

... x = (1 + 4.12)/4 = (5.12)/4 = 1.28

And x = (1 - 4.12)/4 = (3.12)/4 = -0.78
Answer is x= 1.28 OR -0.78

Unless you are specifically directed, you should attempt to use the factorization method before the Formula method.

POINTS TO NOTE

• Care should always be taken in manipulating the negative signs, as this provides the greatest challenge in this method.
• The + or - enables you to obtain two roots

• The entire numerator is over 2a. A common error is to use square root of b² - 4ac over 2a, separating -b. In other words, th eincorrect formula (-b +OR- Square root of b² - 4ac)/2a is sometimes used.
• The value within the square root should always be positive. When this is not so, it usually implies an error in calculation. PLEASE CHECK YOUR WORKING.
• If the value within the square root is negative, then the equation has no real roots.

For Homework, please find the solution of the quadratic equations.

(1) x² - 4x - 8 =0

(2) 2x² + 5x = 9

(3) 3x² - 10x + 5 =0

(4) 2x² - 3x - 4 = 2 - 4x

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