The Question
Solve for x: (x-8)(x+2) ≤ 0
Details
💡 Hint
Start by finding the values of x where the expression equals zero. These ‘critical points’ will help you divide the number line into sections to test.
📝 Solution Steps
- Find the critical values by setting the quadratic expression equal to zero.
- Plot these critical values on a number line.
- Choose a test value from each interval created by the critical values.
- Substitute each test value into the original inequality to determine which intervals satisfy it.
- Write the solution in interval notation or inequality form, remembering to include the critical values if the inequality includes ‘equal to’.
📚 Explanation
This is a quadratic inequality. The first step is to find the ‘critical values’ by setting the expression equal to zero: (x-8)(x+2) = 0. This gives us x=8 and x=-2. These critical values divide the number line into three intervals: x < -2, -2 ≤ x ≤ 8, and x > 8. You can test a value from each interval in the original inequality to see where it holds true. For example, if x=0 (in the middle interval), (0-8)(0+2) = -8 * 2 = -16, which is ≤ 0. This interval works! Alternatively, visualize the graph of y = (x-8)(x+2). It’s a parabola opening upwards, intersecting the x-axis at -2 and 8. The inequality (x-8)(x+2) ≤ 0 asks for where the parabola is below or on the x-axis. This occurs between and including the roots, so the solution is -2 ≤ x ≤ 8.
✅ Answer
-2 ≤ x ≤ 8
⚠️ Common Mistakes
- Incorrectly determining the intervals or the direction of the inequality.
- Sign errors when testing points in the inequality.
- Forgetting to include the critical values in the solution when the inequality is ‘less than or equal to’ (≤) or ‘greater than or equal to’ (≥).