Quadratic Equations

Notes

General form

The general form of a quadratic equation is ax² + bx + c, where a, b, and c are constants and a is not zero

The roots of the quadratic equation is given by

b² - 4ac is called the discriminant of the expression ax² + bx + c

Types of roots

1. if b² - 4ac > 0 - the two roots are real and distinct

2. if b² - 4ac < 0 - the two roots are imaginary

3. if b² - 4ac = 0 - the two roots are real and equal

Roots of equation

Some common relationships:

Range of values of a quadratic function

The quadratic function f(x)= ax² + bx + c has a minimum value if a is positive. (smiley face)

It has a maximum value if a is negative. (sad face)

Range of values of a quadratic function

Case A

Case B

Note: if ax² + bx + c, f(x) < 0 --> this refers to the part of the graph below the x-axis and the corresponding range of values of x that can be found.

If ax² + bx + c, f(x) > 0 --> this refers to the part of the graph above the x-axis

Case C

Case D

Example 1

Example 2

Steps for finding range of values

eg Find the range of values of x for which x² - 5x + 6 < 0

1. Determine max or min curve

if coefficient of x² is positive --> min curve

if coefficient of x² is negative --> max curve

In this case, it is a min curve (curve upwards) because the coefficient of x² is positive.

2. Factorize quadratic equation x² - 5x + 6 into (x - 3)(x - 2) so

(x - 3)(x - 2) = 0 when x = 3, or 2 --> means the curve will cut the x-axis at these 2 points

3. Find the point on the y axis where x = 0 by substituting x = 0 into the equation

hence y = 6

4. Sketch the curve

5. Since we want the range of values x² - 5x + 6 < 0 which is below the x-axis,

the range is 2 < x < 3

Note: if we are looking for x² - 5x + 6 > 0 instead, x < -2 or x > 3

(look at the part above the x-axis)

Note: Below x-axis --> negative values of y (y < 0)

Above x-axis --> positive values of y (y > 0)

Questions

Answers

1. 3/2, 2; 8x² - 18x + 13 = 0

2a. x < -3 or x > 2/3

3a. x < -5 or x > 3

3b. x > 2/3 or x > 3/2

3c. 1 < x < 3/2 or x > 2

3d. x < -4; -1 < x < 2; x > 3

4. -q/p, r/p, -(q/r³)(q² - 3pr)

5. a= -1/5, alpha= -1/2, beta= 1/2 ; a= 3, alpha= 3/2, beta= 5/2

6. (a + 2c)²/ac

7a. 4x² - 32x + 69 = 0

7b. 9x² - 24x + 36 = 0

7c. 9x² - 2x + 9 = 0

8a. x > 2, x < -3

8b. 1/2 < x < 3

9. Let the equation be y

(2y - 1)x² + (3 - 3y)x + 2y - 1 = 0

D = (3 - 3y)² - 4(2y - 1)(2y - 1) > 0

as x is real. That is (7y - 5)(y + 1) < 0

--> -1 < y < 5/7

10a. k < 7

10b. k = 9 or 31.5