# need help

Solve each equation or inequality.  Graph the solution on a number line.  If  there is no solution, write "no solution.

1)2 ≤ |x – 2| ≤ 8

2)x^2– 3x – 18 > 0

and

3) x^2-x-12/x-5 ≤0

(1).

The compound inequality 2 ≤ | x - 2 | ≤ 8.

Write the compound inequality using the word "and". Then solve each inequality.

2 ≤ | x - 2 | and | x - 2 | ≤ 8.

Solve the inequality 1 : 2 ≤ | x - 2 | for x.

| x - 2 | ≥ 2 is equivalent to x - 2 ≥ 2 or x - 2 ≤ - 2. Solve the inequality.

Case 1 : x - 2 ≥ 2 ⇒ x ≥ 4.

Case 2 : x - 2 ≤ - 2 ⇒ x ≤ 0.

Solution of the inequality 2 ≤ | x - 2 | is x ≥ 4 or x ≤ 0.

The interval notation form of solution set is (-∞, 0] U [4, ∞).

Solve the inequality 2 : | x - 2 | ≤ 8 for x.

| x - 2 | ≤ 8 is equivalent to - 8 ≤ x - 2 ≤ 8.

Solve the inequality for x.

Add 2 to each part of the inequality.

- 8 + 2 ≤ x - 2 + 2 ≤ 8 + 2

- 6 ≤ x ≤ 10

Solution of the inequality | x - 2 | ≤ 8 is - 6 ≤ x ≤ 10.

The interval notation form of solution set is [-6, 10].

The combine solution of (-∞, 0] U [4, ∞) and [-6, 10] is the original solution.

Test intervals   x - value        Compound Inequality                  Conclusion

(-∞, 0]              x = - 1           2 ≤ | (-1) - 2 | ≤ 8 ⇒ 2 ≤ 3 ≤ 8         True

x = - 6           2 ≤ | (-6) - 2 | ≤ 8 ⇒ 2 ≤ 8 ≤ 8         True

x = - 7           2 ≤ | (-7) - 2 | ≤ 8 ⇒ 2 ≤ 9 ≤ 8         False

[4, ∞)               x = 5             2 ≤ | (5) - 2 | ≤ 8 ⇒ 2 ≤ 3 ≤ 8           True

x = 10           2 ≤ | (10) - 2 | ≤ 8 ⇒ 2 ≤ 8 ≤ 8         True

x = 11           2 ≤ | (11) - 2 | ≤ 8 ⇒ 2 ≤ 9 ≤ 8        False

[-6, 10]             x = - 5          2 ≤ | (-5) - 2 | ≤ 8 ⇒ 2 ≤ 7 ≤ 8         True

x = 0            2 ≤ | (0) - 2 | ≤ 8 ⇒ 2 ≤ 2 ≤ 8           True

x = 1              2 ≤ | (1) - 2 | ≤ 8 ⇒ 2 ≤ 1 ≤ 8           False

x = 4              2 ≤ | (4) - 2 | ≤ 8 ⇒ 2 ≤ 2 ≤ 8           True

The solution of the original compound inequality is [ -6, 0] U [4, 10] and its graph is

edited Aug 27, 2014 by bradely

(2).

The inequality is x2 - 3x - 18 > 0.

Write the polynomial as factor form

x2 - 3x - 18 = x2 - 6x + 3x - 18

= x(x - 6) + 3(x - 6)

= (x + 3)(x - 6).

The key numbers are x = - 3 and x = 6. So, the polynomial’s test intervals are (-∞, -3), (-3, 6) and (6, ∞).

In each test interval, choose a representative x-value and evaluate the polynomial.

Test Interval  x-value                Polynomial Value                        Conclusion

(-∞, -3)            x = - 4      (-4)2 - 3(-4) - 18 = 16 + 12 - 18 = 10 > 0      Positive

(-3, 6)             x = 0        (0)2 - 3(0) - 18 = 0 - 0 - 18 = - 18 < 0            Negative

(6, ∞)              x = 7        (7)2 - 3(7) - 18 = 49 - 21 - 18 = 10 > 0          Positive

From this you can conclude that the inequality is satisfied on the open intervals (-∞, -3) and (6, ∞). So, the solution set is (-∞, -3) U (6, ∞) and its graph is

(3).

The rational inequality is (x2 - x - 12) / (x - 5) 0.

The zeros (the x-values for which its numerator is zero) and its undefined values (the x-values for which its denominator is zero) of the rational expression are called key numbers.

Write the numerator as factor form

x2 - x - 12 = x2 - 4x + 3x - 12

= x(x - 4) + 3(x - 4)

= (x + 3)(x - 4).

The key numbers are x = - 3, x = 4 and x = 5. So, the polynomial’s test intervals are (-∞, -3), (-3, 4), (4, 5) and (5, ∞).

In each test interval, choose a representative x-value and evaluate the polynomial.

Test Interval  x-value                Polynomial Value                       Conclusion

(-∞, -3)            x = - 4    [(-4)2 - (-4) - 12] / [(-4) - 5] = - 8/9 < 0       Negative

(-3, 4)              x = 0       [(0)2 - (0) - 12] / [(0) - 5] = 12/5 > 0           Positive

(4, 5)               x = 4.5    [(4.5)2 - (4.5) - 12] / [(4.5) - 5] = - 7.5 < 0   Negative

(5, ∞)              x = 6       [(6)2 - (6) - 12] / [(6) - 5]= 18 > 0               Positive

From this you can conclude that the inequality is satisfied on the open intervals (-∞, -3) and (4, 5). Moreover, because (x2 - x - 12) / (x - 5) = 0 when x = -3 and x = 4. So, the solution set is (-∞, -3] U [4, 5) and its graph is