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Solving Inequalities: An Overview (page 3 of 3)

Sections: Linear inequalities, Quadratic inequalities, Other inequalities

General Polynomial Inequalities

• Solve x⁵ + 3x⁴ – 23x³ – 51x² + 94x + 120 > 0. 

First, we factor to find the zeroes:

x⁵ + 3x⁴ – 23x³ – 51x² + 94x + 120

= (x + 5)(x + 3)(x + 1)(x – 2)(x – 4) = 0

...so x = –5, –3, –1, 2, and 4. (Review how to solve polynomials, if you're not sure how to get this solution.)

To solve by the Test-Point Method, pick a sample point in each interval, the intervals being (negative infinity, –5), (–5, –3), (–3, –1), (–1, 2), (2, 4), and (4, positive infinity). As you can see, if your polynomial or rational function has many factors, the Test-Point Method can become quite time-consuming.

To solve by the Factor Method, solve each factor for its positivity: x + 5 > 0 for x > –5; x + 3 > 0 for x > –3; x + 1 > 0 for x > –1; x – 2 > 0 for x > 2; and x – 4 > 0 for x > 4. Then draw the grid:

...and fill it in:

...and solve:

Then the solution (remembering to include the endpoints, because this is an "or equal to" inequality) is all the x-values in the intervals [–5, –3], [–1, 2], and [4, positive infinity]. 

As you can see, if your polynomial or rational function has many factors, the Factor Method can be much faster.

Rational Inequalities

• Solve x/(x – 3) < 2. 

First off, you have to remember that you can't begin solving until you have the inequality in "= 0" format. Here's how the problem works:

Now convert to a common denominator:

...and simplify:   Copyright © Elizabeth Stapel 2006-2008 All Rights Reserved

The two factors are –x + 6 and x – 3. Note that x cannot equal 3, or else you would be dividing by zero, which is not allowed. The first factor, –x + 6, equals zero when x = 6. The other factor, x – 3, equals zero when x = 3. Now, x cannot actually equal 3, so this endpoint will not be included in any solution interval (even though this is an "or equal to" inequality), but we need the value in order to figure out what our intervals are. In this case, our intervals are (negative infinity, 3), (3, 6], and [6, positive infinity). Note the use of brackets to indicate that 6 can be included in the solution, but that 3 cannot.

Using the Test-Point Method, pick a point in each interval and test for the sign on the result. You could use, say, x = 0, x = 4, and x = 7.

Using the Factor Method, solve each factor: –x + 6 > 0 for –x > –6, or x < 6; x – 3 > 0 for x > 3. Do the grid:

...fill in the signs on the factors:

...and solve for the sign on the rational function:

So the solution is all x's in the intervals (negative infinity, 3) and [6, positive infinity).

There is another way to solve inequalities. You still have to find the zeroes (x-intercepts), but then graph the function, and just look: wherever the graph is above the x-axis, the function is positive; wherever it is below the axis, the function is negative. For instance, for the first quadratic we did, y = x² – 3x + 2 > 0, we already know that the zeroes are at x = 1 and x = 2.. Here's the graph:

Then the solution is obvious: take the two intervals (but not the interval endpoints) where the line is above the x-axis.

Or that huge polynomial we did: x⁵ + 3x⁴ – 23x³ – 51x² + 94x + 120 > 0. We already know that the zeroes of the polynomial are at x = –5, x = –3, x = –1, x = 2, and x = 4. Here's the graph:

Then the solution is obvious: take the three intervals (together with the interval endpoints) where the line is above the x-axis.

As you can probably guess, a graphing calculator can save you a lot of time on these inequalities, if you understand what you're doing. You should still show your work, though.

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