# Francesco Giannino » 3.Polynomial inequalities. Rational inequalities. Absolute-value inequalities

### Polynomial inequalities

In mathematics a polynomial inequality can be written as follows:

P(x) ≥ 0, (≤ )

where P(x) is a reducible polynomial.

Example. P(x)= x2x – 6 ≥ 0

### Polynomial inequalities

We will solve polynomial inequalities using the following steps:
Rewrite the inequality so that there is a zero on the right side (for example P (x) > 0).
Find all polynomial factors of the polynomial function on the right side: P(x) = P1(x) · P2(x) ·… · Pn(x) > 0
To find the critical values of polynomial function on the right side, set each polynomial factor to zero and solve for x.
Determine the sign of each polynomial factor in the intervals delimited by its roots.
The final solution includes intervals, given by the rule sign, in which the polynomial function P(x) has the sign satisfying the assigned inequality.

### Exercises

Exercise 6. Solve the following inequality:

x2 – 3 > 2x

Solution. Rewrite the inequality with one side equal to zero

x2 – 2x –3 > 0

Find the zeros of the polynomial x2 – 2x – 3 and factor the left side of the inequality:

(x + 1)(x – 3) > 0

The two real zeros -1 and 3 of the left side of the inequality, divide the real number line into 3 intervals:
(-∞, -1), (-1, 3), (3,+ ∞)

### Exercises

Check the sign of the product (x + 1)(x – 3) on these intervals:

x + 1 > 0 if x > -1
x – 3 > 0 if x > 3

then

Sign of x + 1 > 0
Sign of x – 3 > 0
Sign of (x + 1)(x – 3) > 0
In fact
x + 1 > 0 ⇒  x > -1
x – 3 > 0 ⇒ x > 3
And then (x + 1)(x – 3) > 0 if $\begin{cases} x + 1 > 0 \ \ \text{AND}\ \ x - 3 > 0\\ OR\\ x + 1 < 0 \ \ \text{AND}\ \ x - 3 < 0\\ \end{cases}$

### Exercises

Finally, the solution in interval notation is
(-∞, -1) ∪ (3,+ ∞)
The graph of solution is sketched below.

### Exercises

Exercise 7. Solve the following inequality: − x2 + 4 < 0

Solution. Multiply both sides by -1 and REVERSE (-1 is negative) the inequality sign

x2 – 4 > 0

Find the zeros of the polynomial x2 – 4 and factor the left side of the inequality:

(x – 2)(x +2) > 0

The two real zeros -2 and +2 of the left side of the inequality, divide the real number line into 3 intervals:

(-∞, -2), (-2, 2), (2,+ ∞)

### Exercises

Check the sign of the product (x – 2)(x + 2) on these intervals:

x + 2 > 0 if x > –2
x – 2 > 0 if x > 2

then
Sign of x + 2 > 0
Sign of x – 2 > 0
Sign of (x + 2)(x – 2) > 0
In fact
x – 2 > 0 ⇒ x > 2
x + 2 > 0 ⇒ x > –2
And then (x – 2)(x + 3) > 0 if $\begin{cases} x - 2 > 0 \ \ \text{AND}\ \ x + 2 > 0\\ OR\\ x - 2 < 0 \ \ \text{AND}\ \ x + 2 < 0\\ \end{cases}$

### Exercises

Finally, the solution in interval notation is

(-∞, -2) ∪ (2,+ ∞)

The graph of solution is sketched below.

### Exercises

Exercise 8. Solve the following inequality: x (x2 + 4) > 0
Solution. Let us note that the polynomial factor (x2 + 2) is always positive and the sign of the assigned polynomial depends only on the zero 0 of polynomial factor x. The real zero 0 divides the real number line into two intervals:
(– ∞, 0), (0, + ∞ )

Check the sign of x on these two intervals.

### Exercises

Finally, the solution in interval notation is

(0,+ ∞)

The graph of solution is sketched below.

### Exercises

Exercise 9. Solve the following inequality:

2x2 + 4xx2x – 6.

Solution. Verify that the solution in interval notation is

(-∞, – 3] ∪  [- 2,+ ∞)

Exercise 10. Solve the following inequality:

x2 + 2x – 8 < 0.

Solution. Verify that the solution in interval notation is

[- 4, 2 ]

### Rational inequalities

A rational inequality can be written as follows:

P(x)/Q(x) > 0 (or P(x)/Q(x) > 0)

where P and Q are polynomials.

Example. P(x)/Q(x) = (x – 2) / (x + 3)

### Rational inequalities

The process for solving rational inequalities is nearly identical to the process for solving polynomial inequalities with a few minor differences. The sign of the rational expression P/Q depends on the signs of P and Q. In turn the signs of P and Q depend on the zeros of P and Q respectively if there are any. Hence the sign of P/Q depends on the zeros of both P and Q and changes (if it does!) only at these zeros. So to solve a rational inequality of the form P/Q > 0 (or P/Q < 0), we first find the zeros of P and Q then make a table of sign of P/Q.

### Exercises

Exercise 11. Solve the following rational inequality: $\frac{(x+3)(x-5)}{3(x-1)} > 0$

Solution. The critical numbers for a rational inequality are all the zeros of the numerator and the denominator. Since the numerator and denominator are already factored in this example, we see that the critical numbers are: -3, 5, and 1.
The three critical numbers divide the number line into four intervals:

(-∞, -3), (-3, 1), (1,+ ∞) 1

### Exercises

Check the sign of the rational function on these intervals:

x + 3 > 0 if x > – 3
x – 5 > 0 if x > 5
x – 1 > 0 if x > 1

then

Sign of x + 3 > 0
Sign of x – 5 > 0
Sign of x – 1 > 0
Sign of $\frac{(x+3)(x-5)}{3(x-1)} > 0$

### Exercises

And then the rational function $\frac{(x+3)(x-5)}{3(x-1)} > 0$

is positive if: – 3 < x < 1 and x > 5.
Finally, the solution in interval notation is (- 3, 1) ∪ (5,+ ∞)

The graph of solution is sketched below.

### Absolute-Value Inequalities

Inequalities involving absolute values can be rewritten as combinations of inequalities.

Let a be a positive number.

| x | < a if and only if – a < x < a.

| x | > a if and only if x < – a or x > a.

### Exercises

Exercise 12. Solve the following Inequality involving absolute value:

| x + 2 | < 3

Solution. The above inequality is solved by writing a double inequality equivalent to the given inequality but without absolute value:

– 3 < x + 2 < 3

Solve the double inequality to obtain: – 5 < x < 1
The above solution set is written in interval form as follows: (– 5, 1)
and the graph of solution is sketched below.

### Exercises

Exercise 13. Solve the following Inequality involving absolute value:

| – 2x – 4 | ≥ 9

Solution. Solving the above inequality is equivalent to solving

– 2x – 4 ≥ 9 or – 2x – 4 ≤ – 9

which gives: – 2x ≥ 13 or – 2x ≤ – 5

x ≤ – 13 / 2 or x ≥ 5 / 2

The above solution set is written in interval form as follows:

] – ∞ , – 13 / 2] ∪ [5 / 2 , + ∞ [

and the graph of solution is sketched below.

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