Math 9

2. ≤ ≥
ARE YOU FAMILIAR WITH THESE SYMBOLS?

3. INEQUALITY SYMBOLS:
< = IS LESS THAN
> = IS GREATER THAN
≤ = IS LESS THAN OR EQUAL
≥ = IS GREATER THAN OR EQUAL

4. QUADRATIC INEQUALITY
It is an inequality that contains a polynomial of degree
2 and can be written in any of the following forms.
𝑎𝑥2
+ 𝑏𝑥 + 𝑐 > 0
𝑎𝑥2
+ 𝑏𝑥 + 𝑐 < 0
𝑎𝑥2
+ 𝑏𝑥 + 𝑐 ≥ 0
𝑎𝑥2
+ 𝑏𝑥 + 𝑐 ≤ 0
Where a, b, and c are real numbers and a ≠0.

5. Review
Interval Notation in the Number Line.
X > 4
X< -3
X ≤ -1
X ≥ 5

6. Review
Writing Interval Notation for each Region in the
Number Line.
A B C
(−∞, −𝟒) (−𝟒, 𝟑) (𝟑, +∞)
HOLLOW/ OPEN CIRCLE MEANING THE NUMBERS
WITHIN THE CIRCLE ARE NOT PART OF THE SOLUTION.
USE PARENTHESIS FOR THE INTERVAL NOTATION.
AM I HOLLOW
OR SOLID
CIRCLE?

7. Review
Writing Interval Notation for each Region in the
Number Line.
A B C
(−∞, −𝟏] [−𝟏, 𝟐] [𝟐, +∞)
SOLID/SHADED CIRCLE MEANING THE NUMBERS
WITHIN THE CIRCLE ARE PART OF THE SOLUTION.
USE BRACKET FOR THE INTERVAL NOTATION.
AM I HOLLOW
OR SOLID
CIRCLE?

8. INEQUALITY SYMBOLS:
< = IS LESS THAN
> = IS GREATER THAN
≤ = IS LESS THAN OR EQUAL
≥ = IS GREATER THAN OR EQUAL
OPEN/HOLLOW
CIRCLE
SHADED/SOLID
CIRCLE

9. SOLVING QUDRATIC INEQUALITY
STEP 1: Express the quadratic inequality as a quadratic
equation in the form of 𝑎𝑥2
+ 𝑏𝑥 + 𝑐 = 0 and then solve for x
or the roots of the equation.
STEP 2: Locate the numbers found in step one on number line.
The number line will be divided in 3 regions.
STEP 3: Choose a test point each region and substitute the test
point to the original inequality. If its hold true, then the region
belongs to the solution set, otherwise, it is not part of the
solution set.

10. Example 1:
A B C
𝑥2
− 6𝑥 + 𝟓 < 𝟎
𝑥2 − 6𝑥 + 𝟓 = 𝟎
𝑥 − 1 𝑥 − 5 = 0
𝑥 − 1 = 0 𝑥 − 5 = 0
𝑥 = 1 𝑥 = 5
STEP 1:
STEP 2:

11. STEP 3:
REGION TETSING POINT SOLUTION RESULT
A x = 0 𝑥2
− 6𝑥 + 𝟓 < 𝟎
(0)2 – 6(0) + 5 <0
0 – 0 + 5 < 0
5 < 0
FALSE
B x = 2 𝑥2 − 6𝑥 + 𝟓 < 𝟎
(2)2 – 6(2) + 5 <0
4 – 12 + 5 < 0
-3 < 0
TRUE
C x = 6 𝑥2
− 6𝑥 + 𝟓 < 𝟎
(6)2 – 6(6) + 5 <0
36 – 36 + 5 < 0
5 < 0
FALSE

12. A B C
REGION TETSING POINT SOLUTION RESULT
A x = 0 𝑥2 − 6𝑥 + 𝟓 < 𝟎
(0)2 – 6(0) + 5 <0
0 – 0 + 5 < 0
5 < 0
FALSE
B x = 2 𝑥2 − 6𝑥 + 𝟓 < 𝟎
(2)2 – 6(2) + 5 <0
4 – 12 + 5 < 0
-3 < 0
TRUE
C x = 6 𝑥2 − 6𝑥 + 𝟓 < 𝟎
(6)2 – 6(6) + 5 <0
36 – 36 + 5 < 0
5 < 0
FALSE
(1, 5)
SOLUTION SET:

13. Example 2: 𝑥2
− 2𝑥 ≥ 𝟖