Chapter Closure
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Exercise 138 Page 575
a
How can you isolate x?
b
Create an or
compound inequality because the absolute value is greater than the given value.
c
Try to rewrite this inequality as a compound inequality.
d
Solve the related quadratic equation, plot the solutions on a number line, and test a value from each interval.
a 
Inequality: x≥ 4
Number Line:
b 
Inequality: x< - 1 or x>5
Number Line:
c 
Inequality: - 2 < x < 3
Number Line:
d 
Inequality: - 1 ≤ x ≤ 4
Number Line:
a
Inequalities can be solved in the same way as equations, by performing inverse operations on both sides until the variable is isolated. The only difference is that when you divide or multiply by a negative number, you must reverse the inequality sign.
This inequality tells us that all values greater than or equal to 4 will satisfy the inequality. Below we demonstrate the inequality by graphing the solution set on a number line. Notice that x can equal 4, which we show with a closed circle on the number line.
b
We are asked to solve the given absolute value inequality.
|x-2|>3
To do this we will create a compound inequality by removing the absolute value. In this case the solution set contains the numbers that make the distance between x and 2 greater than 3 in the positive direction or in the negative direction.
Case 1
This inequality tells us that all values greater than 5 will satisfy the inequality.
Case 2
This inequality tells us that all values less than - 1 will satisfy the inequality.
Solution Set
The solution to this type of compound inequality is the combination of the solution sets. First Solution Set:& x>5 Second Solution Set:& x< - 1 Combined Solution Set:& x< - 1 or x>5
Graph
The graph of this inequality includes all values less than - 1 or greater than 5. We show this by keeping the endpoints open.
c
We are asked to solve the given inequality.
|2x-1| < 5
To do this we will create a compound inequality by removing the absolute value. In this case, the solution set is any number less than 5 away from the midpoint in the positive direction and any number less than 5 away from the midpoint in the negative direction.
Case 1
This inequality tells us that all values less than 3 will satisfy the inequality.
Case 2
This inequality tells us that all values greater than - 2 will satisfy the inequality.
Solution Set
The solution to this type of compound inequality is the overlap of the solution sets. Let's recombine our cases back into one compound inequality. First Solution Set:& x < 3 Second Solution Set:& - 2 < x Intersecting Solution Set:& - 2 < x < 3
Graph
The graph of this inequality includes all values from - 2 to 3, not inclusive. We show this by using open circles on the endpoints.
d
To solve the quadratic inequality algebraically, we will follow three steps.
- Solve the related quadratic equation.
- Plot the solutions on a number line.
- Test a value from each interval to see if it satisfies the original inequality.
Step 1
We will start by solving the related equation.
x=- b±sqrt(b^2-4ac)/2a
SubstituteValues
Substitute values
x=- ( - 3)±sqrt(( - 3)^2-4( 1)( - 4))/2( 1)
▼
Simplify right-hand side
NegNeg
- (- a)=a
x=3±sqrt((- 3)^2-4(1)(- 4))/2(1)
CalcPow
Calculate power
x=3±sqrt(9-4(1)(- 4))/2(1)
IdPropMult
Identity Property of Multiplication
x=3±sqrt(9-4(- 4))/2
MultNegNegOnePar
- a(- b)=a* b
x=3±sqrt(9+16)/2
AddTerms
Add terms
x=3±sqrt(25)/2
CalcRoot
Calculate root
x=3 ± 5/2
Now we can calculate the first root using the positive sign and the second root using the negative sign.
| x=3 ± 5/2 | |
|---|---|
| x_1=3 + 5/2 | x_2=3 - 5/2 |
| x_1=8/2 | x_2=- 2/2 |
| x_1=4 | x_2=- 1 |
Step 2
The solutions of the related equation are - 1 and 4. Let's plot them on a number line. Since the original is not a strict inequality, the points will be closed.
Step 3
Finally, we must test a value from each interval to see if it satisfies the original inequality. Let's choose a value from the first interval, x ≤ - 1. For simplicity, we will choose x=- 2.
x^2-3x-4 ≤ 0
Substitute
x= - 2
( - 2)^2 -3( - 2)-4? ≤0
▼
Simplify left-hand side
CalcPow
Calculate power
4-3(- 2)-4? ≤0
MultNegNegOnePar
- a(- b)=a* b
4+6-4? ≤0
AddSubTerms
Add and subtract terms
6 ≰ 0
Since x=- 2 did not produce a true statement, the interval x ≤ - 1 is not part of the solution. Similarly, we can test the other two intervals.
| Interval | Test Value | Statement | Is It Part of the Solution? |
|---|---|---|---|
| - 1 ≤ x ≤ 4 | 0 | - 4 ≤ 0 ✓ | Yes |
| x ≥ 4 | 5 | 6 ≰ 0 * | No |
We can now write the solution and show it on a number line. - 1 ≤ x ≤ 4
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