Inequalities (Part 2) 1.6(2) How to solve inequalities with a degree greater than 1

First, set one side equal to 0. Then, factor into linear and irreducible quadratic factors and find zeros. Try that process with this inequality.

First, set one side equal to 0. Then, factor into linear and irreducible quadratic factors and find zeros. Try that process with this inequality. 2x 2 – x – 3 < 0 What are the zeros?

Next, locate the zeros on a number line and construct a sign chart above it. Fill in the chart. In what interval is the overall sign negative? That is what we want. Notice how the sign stays consistent within one interval. Sign of (2x - 3) Sign of (x + 1) Overall sign

Next, locate the zeros on a number line and construct a sign chart above it. Fill in the chart. In what interval is the overall sign negative? That is what we want. How would you graph to check your answer? Sign of (2x - 3) Sign of (x + 1) Overall sign + - +

Graph to corroborate your work Graph y = 2x 2 – x – 3. Where are the y values negative? Does this match our earlier work?

Warning, Will Robinson! Danger! Factoring inequalities leads to different types of solutions than factoring equalities. For example, (2x-3)(x+1)=0 leads to the solutions x= 3/2 and -1. But (2x-3)(x+1)<0 does not lead to (2x-3)<0 and (x+1)<0. Rather, < leads to a situation in which either one factor or the other is negative at a time.

The next two examples Are on the CAS document I’m sending out. Find the solution set for each one using the same three methods we just used. We’ll talk as a class at the end of each one.

The process can be simpler if the inequality is already factored, but even then you have to stay on your toes. What are the zeros of this rational expression? What to do with the quadratic factor? Now, build the chart.

The process can be simpler if the inequality is already factored, but even then you have to stay on your toes. What are the zeros of this rational expression? What is the solution set in interval notation? (x+2) (3-x) (x+1) Overall

The process can be simpler if the inequality is already factored, but even then you have to stay on your toes. What are the zeros of this rational expression? (x+2) (3-x) (x+1) Overall

The process can be simpler if the inequality is already factored, but even then you have to stay on your toes. What are the zeros of this rational expression? What is the solution set in interval notation? Why the closed intervals? Graph on CAS to verify the solution.

The process can be simpler if the inequality is already factored, but even then you have to stay on your toes. What are the zeros of this rational expression?

The process can be simpler if the inequality is already factored, but even then you have to stay on your toes. What are the zeros of this rational expression? What is the solution set in interval notation? Solve this on CAS.

Try your creative juices on this one. It looks complicated, but a little analysis and simplification will help. First, factor and reduce. to

Now you’re in a position to finish. Consider the parts. The term equals 0 if (2x+1) = 0, or x = -1/2. The numerator is always positive, so you don’t need to include it in the sign chart. For the rest, the term is always greater than 0 if the denominator is positive. What does the sign chart look like?

Now you’re in a position to finish. Consider the parts. What does the sign chart look like? x (x+1) Overall + - +

Now you’re in a position to finish. Consider the parts. Final answer: x 0, x = - 1/2, x ≠ 1. In interval notation: Graph the original inequality and check.

Now you’re in a position to finish. Consider the parts.

Final answer: x 0, x = - 1/2, x ≠ 1. In interval notation: Solve the original equation with CAS. What message do you get? What does it mean?