- Thread starter
- #1

- Thread starter WannaBe
- Start date

- Thread starter
- #1

- Jan 17, 2013

- 1,667

What have you tried ? how to solve that inequality ?

- Thread starter
- #3

(x+3) (7-x) ((x-2)^2) > 0}What have you tried ? how to solve that inequality ?

x > -3 , x < 7 , x > 2

(-3,2)U(7,oo)

Is it correct?

- Jan 17, 2013

- 1,667

\(\displaystyle ab>0 \,\,\, \text{iff }\,\,\,\, a>0,b>0 \,\,\, \text{or}\,\,\, a<0,b<0\)

- Thread starter
- #5

So, whats the next step?

\(\displaystyle ab>0 \,\,\, \text{iff }\,\,\,\, a>0,b>0 \,\,\, \text{or}\,\,\, a<0,b<0\)

- Jan 17, 2013

- 1,667

Note that if you use numbers greater than 7 the inequality will not hold.

- Admin
- #7

I don't want to trample on the help being given by

Suppose we are given the set:

\(\displaystyle \{x|(x+4)(x-1)(x-5)^2(x-12)^3>0\}\)

Step 1: Draw a real number line and mark the roots of the polynomial expression:

Step 2: Consider whether the inequality is weak or strict. If weak, put solid dots at the roots to show they are part of the solution set (giving closed intervals), and if strict put hollow dots to indicate they are not part of the solution (giving open intervals). For this problem, we have a strict inequality so we will put hollow dots at the roots:

Step 3: Choose test values from within each interval into which the roots have divided the real number line. I will choose some and put them in red, however the choice of values is up to the person working the problem, as long as they are within the intervals:

Step 4: Put each test value into each factor of the polynomial, and record the resulting signs of each factor, and the consider what the sign the resulting product must be. An even number of negatives gives a positive while an odd number of negatives gives a negative. Take care to make sure the factors with exponents are counted the correct number of times:

As you become more proficient at this step, you will see that when a root has an odd multiplicity, the sign of the polynomial will change across the root, and when the root has an even multiplicity it will not. The multiplicity of a root refers to how many times it occurs, as indicated by its exponent. Notice the root $x=5$ is of multiplicity 2, and the sign of the polynomial did not change, whereas the root $x=12$ is of multiplicity 3 and the sign did change, as it did also for the other roots of multiplicity 1. Using this information, it is then really only necessary to check one interval, and then apply the information regarding the multiplicity of each root accordingly.

Step 5: Since we are interested in those intervals where the polynomial is positive, we then shade those intervals which resulted in a positive sign:

Step 6: Write the solution in interval notation:

\(\displaystyle (-4,1)\,\cup\,(12,\infty)\)