NotesGeneral formThe general form of a quadratic equation is ax^{2} + bx + c, where a, b, and c are constants and a is not zero
The roots of the quadratic equation is given by b^{2}  4ac is called the discriminant of the expression ax^{2} + bx + c
Types of roots1. if b^{2 } 4ac > 0  the two roots are real and distinct
2. if b^{2}  4ac < 0  the two roots are imaginary
3. if b^{2}  4ac = 0  the two roots are real and equal
Roots of equationSome common relationships:
Range of values of a quadratic functionThe quadratic function f(x)= ax^{2} + 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 ACase B Note: if ax^{2} + bx + c, f(x) < 0 > this refers to the part of the graph below the xaxis and the corresponding range of values of x that can be found. If ax^{2} + bx + c, f(x) > 0 > this refers to the part of the graph above the xaxis
Case C
Case D Example 1
Example 2
Steps for finding range of valueseg Find the range of values of x for which x^{2}  5x + 6 < 0
1. Determine max or min curve if coefficient of x^{2} is positive > min curve if coefficient of x^{2} is negative > max curve
In this case, it is a min curve (curve upwards) because the coefficient of x^{2 }is positive. 2. Factorize quadratic equation x^{2}  5x + 6 into (x  3)(x  2) so (x  3)(x  2) = 0 when x = 3, or 2 > means the curve will cut the xaxis 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 x^{2}  5x + 6 < 0 which is below the xaxis, the range is 2 < x < 3
Note: if we are looking for x^{2}  5x + 6 > 0 instead, x < 2 or x > 3
(look at the part above the xaxis)
Note: Below xaxis > negative values of y (y < 0) Above xaxis > positive values of y (y > 0)
Questions
Answers1. 3/2, 2; 8x^{2}  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/r^{3})(q^{2}  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. 4x^{2}  32x + 69 = 0 7b. 9x^{2}  24x + 36 = 0 7c. 9x^{2}  2x + 9 = 0 8a. x > 2, x < 3 8b. 1/2 < x < 3 9. Let the equation be y (2y  1)x^{2 }+ (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
