Nonlinear Inequalities

A nonlinear inequality involves expressions such as polynomials and rational functions. The standard approach converts the inequality to the form $P > 0$ (or $P < 0$ , $P \geq 0$ , $P \leq 0$ ), factors $P$ , and uses a sign table to identify the solution intervals.

To solve nonlinear inequalities, we follow these steps:

Transfer all terms to one side of the inequality sign and express the inequality in the form

  P
  >
  0
  ,
  
  or
  
  P
  <
  0

or

  P
  ≥
  0
  ,
  
  or
  
  P
  ≤
  0

where 
  P
 is an expression in the variable (usually 
  x
) [to indicate its dependence on 
  x
, we may write it as 
  P
  (
  x
  )
 if we wish]
Factor 
  P


  P
  =
  
    Q
    
      1
    
  
  
    Q
    
      2
    
  
  ⋯
  
    Q
    
      n
    
  

where 
  
    Q
    
      1
    
  
  ,
  …
  ,
  
    Q
    
      n
    
  
 are expressions in 
  x
.
Determine the zeros of each factor of 
  P
 (find the values for which each 
  
    Q
    
      i
    
  
 is zero). These values divide the real line into intervals.
Make a sign diagram (or sign table). Determine the sign of each factor in each interval.
Determine the sign of 
  P
 in each interval using the sign table. Recall that a product (or a quotient) that involves an even number of negative factors is positive and one that involves an odd number of negative factors is negative. If the inequality sign is 
  ≥
 or 
  ≤
, pay attention to the endpoints of the intervals and check if they satisfy the inequality.

For fractions; that is, when $P = \frac{Q_{1}}{Q_{2}}$ we follow the same steps.

How to Determine the Sign of Each Factor

To determine the sign of each factor, we can choose an arbitrary number (called a test value) in that interval and find the sign of the factor. Alternatively you can use the following facts:

If we have a factor of the form $a x + b$

If we have a factor of the form $a x^{2} + b x + c$ , then there are two cases:

Algebraic proof for why the sign of

a x^{2} + b x + c

is the same as

a

if there are no zeros

Given: The quadratic equation $a x^{2} + b x + c = 0$ has no real solutions.
Implication: The discriminant $D = b^{2} - 4 a c$ satisfies $D < 0$ .
Objective: Show that $a x^{2} + b x + c$ is always positive if $a > 0$ , and always negative if $a < 0$ .

Rewrite the Quadratic in Vertex Form: \begin{aligned} ax^2 + bx + c &= a\left(x^2 + \frac{b}{a}x\right) + c \\ &= a\left[\left(x + \frac{b}{2a}\right)^2 - \frac{b^2}{4a^2}\right] + c \\ &= a\left(x + \frac{b}{2a}\right)^2 - \frac{b^2}{4a} + c \\ &= a\left(x + \frac{b}{2a}\right)^2 + \left(c - \frac{b^2}{4a}\right). \end{aligned}
Analyze the Components: The expression a x 2 + b x + c is now expressed as: a ( x + b 2 a ) 2 ⏟ ≥ 0 + ( c − b 2 4 a ) ⏟ Constant .
- The term ${(x + \frac{b}{2 a})}^{2}$ is always non-negative (a square is $\geq 0$ ).
- The constant term $c - \frac{b^{2}}{4 a}$ is influenced by the discriminant $D$ .
Use the Discriminant Condition $D < 0$ : From D = b 2 − 4 a c < 0 , we derive: b^2 < 4ac \quad \Rightarrow \quad \left\{\begin{aligned} &c > \frac{b^2}{4a} \quad \text{if } a > 0, \\ &c < \frac{b^2}{4a} \quad \text{if } a < 0. \end{aligned}\right.
- Case 1: If $a > 0$ , then $c - \frac{b^{2}}{4 a} > 0$ . The quadratic becomes: $a (non-negative term) + positive constant > 0 for all x .$
- Case 2: If $a < 0$ , then $c - \frac{b^{2}}{4 a} < 0$ . The quadratic becomes: $a (non-negative term) + negative constant < 0 for all x .$
Conclusion:
- When $a > 0$ , the expression $a x^{2} + b x + c$ is a sum of non-negative and positive terms, so it is always positive.
- When $a < 0$ , the expression is a sum of non-positive and negative terms, so it is always negative.

Summary: The absence of real roots (

D < 0

) ensures the quadratic expression

a x^{2} + b x + c

never crosses zero. Its sign is governed by the leading coefficient

a

, as shown by completing the square and analyzing the discriminant's constraint. Thus:

Sign (a x^{2} + b x + c) = Sign (a) when D < 0.

Solve $x^{2} + 6 > 5 x$

Solution

x^{2} + 6 > 5 x

x^{2}-5x+6>0\tag{subtract $5x$} (x-2)(x-3)>0\tag{factor} Because the sign of a product depends on the signs of its factors, we need to determine where each factor is positive or negative. We identify that the zeros of these factors are

x = 2

and

x = 3

and using these values we construct the following sign table. Because the product is positive where both factors are positive or both are negative, the solutions are the real numbers in the union

(- \infty, 2) \cup (3, \infty)

{x | x < 2 or x > 3},

as illustrated in the last row of the following sign table.

A sign chart for the quadratic inequality x² − 5x + 6 loading=

Alternatively because the multiple of

x^{2}

(

a = 1

) is positive, as discussed above we know that when

2 < x < 3

, the sign of the polynomial is opposite to the sign of

a

(here negative) and outside of that interval, the sign of the polynomial is the same as the sign of

a

(here positive).

Solve $x^{3} + 2 x^{2} \leq x + 2$

Solution

\begin{aligned} x^{3}+2x^{2} & \le x+2 && \text{(given inequality)}\\ x^{3}+2x^{2}-x-2 & \le0 && \text{(move all terms to one side)} \end{aligned} Now we need to factor

x^{3} + 2 x^{2} - x - 2

. We can do that by grouping terms as follows \begin{aligned} x^{3}+2x^{2}-x-2 & =(x^{3}+2x^{2})-(x+2)\\ & =x^{2}(x+2)-(x+2)\\ & =(x+2)(x^{2}-1) && \text{(factor out } (x+2) \text{)}\\ & =(x+2)(x-1)(x+1) && \text{(use the Difference of Squares formula)} \end{aligned} So we need to find

x

such that

(x + 2) (x - 1) (x + 1) \leq 0

To determine the sign, first we identify the zeros that are

x = - 2

x = - 1

, and

x = 1

, and then construct the following sign table

A sign chart for the polynomial inequality (x + 2)(x − 1)(x + 1) ≤ 0. The number line is divided by the roots -2, -1, and 1. Rows display signs for factors (x + 2), (x − 1), and (x + 1). The final row indicates the product is negative for x < -2, positive for -2 < x < -1, negative for -1 < x < 1, and positive for x loading=

The last row of the sign table is obtained from the fact that the polynomial is the product of the three factors. We read from this table that

(x + 2) (x - 1) (x + 1)

x^{3} + 2 x^{2} - x - 2

is negative or zero on the interval

(- \infty, - 2]

and

[- 1, 1]

. We have included

x = - 2

and

x = \pm 1

because at these points, the polynomial is zero and the inequality is satisfied. The solution set can be expressed as

{x | x \leq - 2 or - 1 \leq x \leq 1} = (- \infty, - 2] \cup [- 1, 1] .

Solve $\frac{x}{x + 1} \leq 2$ .

Solution

Recall that if we multiply both sides by a positive quantity, the direction of the inequality will be preserved but if we multiply both sides by a negative quantity, the direction of the inequality will be reversed. We do not know the sign of

x + 1

, so if we wanted to multiply both sides by

x + 1

, we would not be sure about the direction of the inequality. \begin{aligned} \frac{x}{x+1} & \leq2 && \text{(given inequality)}\\ \frac{x}{x+1}-2 & \leq0 && \text{(subtract 2 from both sides)}\\ \frac{x}{x+1}-\frac{2(x+1)}{x+1} & \leq0 && \text{(common denominator } (x+1) \text{)}\\ \frac{x-2x-2}{x+1} & \le0 && \text{(combine fractions)}\\ \frac{-x-2}{x+1} & \leq0 && \text{(simplify)} \end{aligned} The sign of

(- 2 - x) / (x + 1)

is solely determined by the signs of

(- x - 2)

and

(x + 1)

. To construct the sign table, we need to identify the zeros of the numerator and denominator:

x = - 1

and

x = - 2

. These are the only values at which the fraction may change sign. Now we construct the following sign table.

A sign chart for the rational inequality (-x − 2)/(x + 1) ≤ 0. The number line is divided by critical points -2 and -1. Rows show signs for the numerator -x − 2 and the denominator x + 1. The final row shows the quotient is negative for x < -2, positive for -2 < x < -1, and negative for x loading=

For

x = - 1

, the denominator becomes zero, and because division by zero is not defined, we must eliminate

x = - 1

from the set of acceptable solutions. From the sign table, we read that

(- x - 2) / (x + 1) \leq 0

when

x \leq - 2 or x > - 1

and the solution set is

(- \infty, - 2] \cup (- 1, \infty)

We have included

x = - 2

because for

x = - 2

the numerator and hence the fraction are zero and the inequality is satisfied.

Solve $\frac{x + 1}{x^{2} + 1} > 1$ .

Solution

\begin{aligned} \frac{x+1}{x^{2}+1} & >1 && \text{(given inequality)}\\ \frac{x+1}{x^{2}+1}-1 & >0 && \text{(subtract 1 from both sides)}\\ \frac{x+1}{x^{2}+1}-\underbrace{\frac{x^{2}+1}{x^{2}+1}}_1 & >0 && \text{(common denominator } (x^2+1) \text{)}\\ \frac{x-x^{2}}{x^{2}+1} & >0 && \text{(combine fractions)}\\ \frac{x(1-x)}{x^{2}+1} & >0 && \text{(factor)} \end{aligned} Because

x^{2} + 1 > 0

, the sign of the fraction is the same as the sign of the numerator

x (1 - x)

. The zeros of the numerator are

x = 0

and

x = 1

that divide the real line into three intervals. Then we construct the sign table as follows.

A sign chart for the rational inequality x(1 − x)/(x² + 1) loading=

The sign table shows that

x (1 - x) / (x^{2} + 1) > 0

when

0 < x < 1

. In other words, the solution set is

{x | 0 < x < 1} = (0, 1) .

Because the denominator is never zero, we do not have to exclude any value of

x

Frequently Asked Questions

What is a nonlinear inequality?

A nonlinear inequality is an inequality involving expressions that are not linear, such as polynomials (

x^{2} - 5 x + 6 > 0

) or rational expressions (

\frac{x}{x + 1} \leq 2

). Unlike linear inequalities, the solution is typically a union of intervals found by factoring and using a sign table.

What is a sign diagram (sign table)?

A sign table organizes the sign of each factor across the intervals determined by the zeros of those factors. Each row shows the sign of one factor, and the bottom row combines them to show the sign of the full expression. Where the expression's sign matches the inequality, those intervals form the solution set.

Why do we move all terms to one side before solving a nonlinear inequality?

Moving all terms to one side gives the form

P > 0

P \leq 0

, which lets you analyze the sign of a single expression

P

. Attempting to solve directly (for example, by cross-multiplying a rational inequality without knowing the sign of the multiplied expression) risks reversing the inequality direction unexpectedly.

Why must we exclude points where the denominator is zero?

Division by zero is undefined. Even if the zero of the denominator appears as a boundary point in the sign table, it is never part of the solution set. For a rational inequality

\frac{Q_{1}}{Q_{2}} \leq 0

, the points where

Q_{2} = 0

are always excluded, even when the inequality uses

\leq

\geq

When does a quadratic factor keep the same sign throughout?

A quadratic

a x^{2} + b x + c

keeps the same sign throughout if its discriminant

D = b^{2} - 4 a c < 0

(no real roots). In that case, its sign equals the sign of the leading coefficient

a

: it is always positive if

a > 0

, and always negative if

a < 0

How to Determine the Sign of Each Factor

Frequently Asked Questions

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