PRECALCULUS:GRAPH...-NASTA ED.(FLORIDA)
PRECALCULUS:GRAPH...-NASTA ED.(FLORIDA)
10th Edition
ISBN: 9780136174318
Author: Demana
Publisher: PEARSON
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Chapter P.7, Problem 48E
Solution

The solution set of the absolute value inequality 2x2+3x−20>10 .

  −∞,−4.69∪−3.11,1.61∪3.19,∞

Given:

The absolute value inequality 2x2+3x−20>10 .

Concept used:

  • Absolute value inequality states that, if x<a , then the solutions are −a<x<a .
  • The value of an absolute value function is always greater than or equal to 0.
  • If there is the strict inequality symbol < or> , then use a parenthesis on the interval
  • If there is the inequality symbol ≤ or≥ , then use a bracket on the interval
  • To graph the endpoint with strict inequality < or>, use the open parenthesis or a hollow circle at that endpoint.
  • To graph the endpoint with inequality symbol ≤ or≥, use the bracket or a solid circle at that endpoint.
  • Multiplying or dividing an inequality by a negative number reverse the inequality sign.

Calculation:

In order to solve the given absolute value inequality, use the definition of absolute value function and simplify further as shown below

  2x2+3x−20>102x2+3x−20−10>10−102x2+3x−20−10>0

In order to solve the inequality, plot the graph of the function 2x2+3x−20−10 and locate the intervals where the graph is above the x -axis.

The graph of the function is shown below,

  PRECALCULUS:GRAPH...-NASTA ED.(FLORIDA), Chapter P.7, Problem 48E

Thus, the solution set of the given absolute value inequality is −∞,−4.69∪−3.11,1.61∪3.19,∞ .

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Chapter P.7, Problem 47E
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Chapter P, Problem 1RE

Chapter P Solutions

PRECALCULUS:GRAPH...-NASTA ED.(FLORIDA)

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