Chapter Closure
Sign In
Treat the inequality as if it was an equation and solve for x using the Quadratic Formula. This gives you the boundary point(s) of the inequality.
Find the boundary point(s) by solving the absolute value equation |3x-2|=10.
Solution Set: -3
Diagram:
Solution Set: x≤-8/3 and x≥4
Diagram:
To solve the inequality we must first determine its boundary point(s). To do that, we will treat the inequality as if it was an equation and solve for x.
Inequality:& x^2-2x-15<0
Equality:& x^2-2x-15=0
Let's solve it by using the Quadratic Formula.
Use the Quadratic Formula: a = 1, b= -2, c= -15
- (- a)=a
(- a)^2=a^2
Multiply
Add terms
Calculate root
State solutions
(I), (II): Add and subtract terms
(I), (II): Calculate quotient
The boundary points are located at x=-3 and x=5. Let's mark them on a number line. Notice that the inequality is strict. This means the boundary points are not part of the solution set. To determine where we should shade the number line, we will test three numbers — one in each of the intervals defined by the boundary points.
By substituting x with -5, 0, and 6 in the inequality we can determine in which region the inequality holds true. If the inequality holds true, we should shade this region. Otherwise, we do not shade it. |c|c|c| [-0.8em] -3pt x -3pt & -1pt x^2-2x-15<0 -1pt & -1pt Evaluate -1pt [0.5em] [-1em] -3pt -5 -3pt & -1pt (-5)^2-2(-5)-15? <0 -1pt& -1pt 20 ≮ 0 * -1pt [0.5em] [-1em] -3pt -3pt & ^2-2( )-15? <0 & -4pt -15 < 10 ✓ -1pt [0.5em] [-1em] -3pt 6 -3pt & 6^2-2(6)-15? <0 & 9 ≮ 0 * [0.5em] As we can see, the inequality is true for values between the boundary points. Therefore, we should shade this region.
Like in Part A, we should first find the inequality's boundary point(s). We will do that by temporarily considering the related equation instead.
Inequality:& |3x-2|≥10
Equality:& |3x-2|=10
The absolute value of a number measures the distance from 0 to that number. In this case, the distance from 0 to 3x-2 is 10. The only numbers whose distance to 0 is 10 are 10 and - 10. Therefore, 3x-2 can be either 10 or - 10.
(I), (II): LHS+2=RHS+2
(I), (II): .LHS /3.=.RHS /3.
The boundary points are located at x=4 and x=- 83. Let's mark them on a number line. Notice that the inequality is non-strict. This means the boundary points are part of the solution set. To determine where we should shade the number line, we will test three numbers — one in each of the intervals defined by the boundary points.
Now, we will substitute -4, 0 and 5 for x in the given inequality and check if the inequality is true. |c|c|c| [-0.8em] x & |3x-2| ≥ 10 & Evaluate [0.5em] [-1em] -4 & |3(-4)-2|? ≥ 10 & 14 ≥ 10 ✓ [0.5em] [-1em] & |3( )-2|? ≥ 10 & 2 ≱ 10 * [0.5em] [-1em] 5 & |3(5)-2|? ≥ 10 & 13 ≥ 10 ✓ [0.5em] The inequality is true to the left of the first boundary point and to the right of the second boundary point. Therefore, we will shade these regions.
