1. Quadratic Inequalities with 1 Variable (Interval analysis)
Systems of Equations and Inequalities

4. Process: We use a process called test intervals
1. Find zeros (critical numbers, values making function = 0) to make up your intervals
2. Choose any number from each interval and determine whether it makes the function + or -, the result represents the sign of the entire interval

5. Example: x2 - 2x < 24 Steps to the solution:
1. We first make the equation 0 on one side.
x2 – 2x – 24 < 0
2. Next, we factor to find the critical points.
(x - 6)(x + 4) < 0
This means we want the interval where the function is NEGATIVE.

6. (x + 4) (x – 6) (x+4)(x – 6) (-∞, -4) (-4, 6) (6, ∞)
3. Create a chart with the intervals between the roots down the side and the factors across the top.
(x + 4)
(x – 6)
(x+4)(x – 6)
(-∞, -4)
(-4, 6)
(6, ∞)

7. 4. Test values in each interval to determine the sign of the factor in that interval.
(x + 4)
(x – 6)
(x+4)(x – 6)
(-∞, -4)
(-4, 6)
(6, ∞)
In the first interval, test a number less than -4, for example -5.
In the second interval test a number between -4 and 6, for example 0.
In the third interval, test a number above 6, for example 7.

8. 5. Multiply the signs of the factors in each interval together to determine whether the interval is positive or negative.
(x + 4)
(x – 6)
(x+4)(x – 6)
(-∞, -4)
(-4, 6)
(6, ∞)
So the solution is ___________________________________.

9. Check answer graphically.
We can see that it intersects the x–axis at –4 and 6 and is below the x–axis for that interval.
Therefore, we are correct!

10. Example 2 5x2 – x + 3 > 3x2 – 7x + 11 1. Make equation = 0
Think: where is the function POSITIVE
2. Factor and find roots.
2(x2 + 3x – 4) > 0
2(x + 4) (x – 1) > 0
x = and x = 1

11. 3. Build and fill out chart.

12. Solution:
So now we can see that the function is positive for the intervals x < –4 and x > 1.

13. Example 3 x2 – 4x > 10 x2 – 4x – 10 > 0
(use quadratic formula to solve for x)

14. Chart.

15. Assignment:
Assignment: Pg. 484–485 # 1, 3(a,b), 4, 5, 9, 12, 13