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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
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