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Higher Order Inequalities

These inequalities have variables whose exponents are greater than 1.
 
For example, x2 + 4 < 2 and x3 – 9 > 0. The number line is often helpful in solving these types of inequalities.

 

Example

 For which values of x is x2 > – 6x – 5 ?

 

First, replace the inequality symbol with an equal symbol: x2 = – 6x – 5
Adding 6x and 5 to both sides yields x2 + 6x + 5 = 0
Factoring yields (see General Trinomials in the chapter Factoring) (x + 5)(x + 1) = 0
Setting each factor to 0 yields x + 5 =0 and x + 1 =0
Or x = –5 and x = –1
 
Now, the only numbers at which the expression can change sign are –5 and –1. So –5 and –1 divide the number line into three intervals. Let’s set up a number line and choose test points in each interval:

 

When x = –6, x2 > – 6x – 5 becomes 36 > 31. This is true.

 

Hence, all numbers in Interval I satisfy the inequality. That is, x < –5. When x = –3, x2 > – 6x – 5 becomes 9 > 13. This is false.

 

Hence, no numbers in Interval II satisfy the inequality. When x = 0, x2 > – 6x – 5 becomes 0 > –5. This is true.

 

Hence, all numbers in Interval III satisfy the inequality. That is, x > –1. The graph of the solution follows:

 
 

Note, if the original inequality had included the greater-than-or-equal symbol, >, the solution set would have included both –5 and –1. On the graph, this would have been indicated by filling in the circles above –5 and –1. The open circles indicate that –5 and –1 are not part of the solution.
 

 

Summary of steps for solving higher order inequalities:
  1. Replace the inequality symbol with an equal symbol.
  2. Move all terms to one side of the equation (usually the left side).
  3. Factor the equation.
  4. Set the factors equal to 0 to find zeros.
  5. Choose test points on either side of the zeros.
  6. If a test point satisfies the original inequality, then all numbers in that interval satisfy the inequality. Similarly, if a test point does not satisfy the inequality, then no numbers in that interval satisfy the inequality.





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