Explanation

x² − x − 4 ≤ 2 Subtract two from both sides to rewrite it in the quadratic standard form: = x² − x −4 −2 ≤ 2 −2 = x² − x − 6 ≤ 0 Now set it = 0 and factor and solve like normal. = x² − x - 6 =0 = (x − 3)(x + 2) = 0 x + 2 = 0 or x - 3 = 0 x = -2 or x = 3 So the two zeros are -2 and 3, and will mark the boundaries of our answer interval. To find out if the interval is between -2 and 3, or on either side, we simply take a test point between -2 and 3 (for instance, x = 0) and evaluate the original inequality. = x² − x − 4 ≤ 2 = (0)² − (0) − 4 ≤ 2 = 0 − 0 − 4 ≤ 2 −4 ≤ 2 Since the above is a true statement, we know that the solution interval is between -2 and 3, the same region where we picked our test point. Since the original inequality was less than or equal, we include the endpoints. ∴ −2 ≤ x ≤ 3.

x² − x − 4 ≤ 2 Subtract two from both sides to rewrite it in the quadratic standard form: = x² − x −4 −2 ≤ 2 −2 = x² − x − 6 ≤ 0 Now set it = 0 and factor and solve like normal. = x² − x - 6 =0 = (x − 3)(x + 2) = 0 x + 2 = 0 or x - 3 = 0 x = -2 or x = 3 So the two zeros are -2 and 3, and will mark the boundaries of our answer interval. To find out if the interval is between -2 and 3, or on either side, we simply take a test point between -2 and 3 (for instance, x = 0) and evaluate the original inequality. = x² − x − 4 ≤ 2 = (0)² − (0) − 4 ≤ 2 = 0 − 0 − 4 ≤ 2 −4 ≤ 2 Since the above is a true statement, we know that the solution interval is between -2 and 3, the same region where we picked our test point. Since the original inequality was less than or equal, we include the endpoints. ∴ −2 ≤ x ≤ 3.