Solve x² + 17x + 72 ≥ 0
Quadratic inequality, worked out line by line the way a teacher would write it.
Answer
| Solution | x ≤ −9 or x ≥ −8 |
| Interval notation | (−∞, −9] ∪ [−8, ∞) |
Step-by-step solution
6 steps-
1 Givenx^{2} + 17 x + 72 \ge 0
Solve for x.
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2 Factor\left(x + 8\right) \left(x + 9\right) \ge 0
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3 Critical points: where the expression is 0 or undefined
A polynomial or a fraction can only change sign where it equals 0 or where it is undefined. Between two neighbouring critical points the sign stays the same, so one test value decides each whole interval.
\text{zeros: } x = -8,\; x = -9The sign can only change at these points, so test one value in each interval between them.
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4 Sign chart: the sign of each factor in each interval
Each row shows the sign of one factor in each interval, and the bottom row multiplies them: an even number of minus signs gives +, an odd number gives −.
\begin{array}{c|ccccc} & \left(-\infty,\, -9\right) & -9 & \left(-9,\, -8\right) & -8 & \left(-8,\, \infty\right) \\ \hline x + 8 & - & - & - & 0 & + \\ \hline x + 9 & - & 0 & + & + & + \\ \hline \text{whole} & + & 0 & - & 0 & + \end{array}Test values: x = -10, x = -17/2, x = -7.
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5 Keep the intervals where the expression is positive
Keep the intervals whose sign matches the inequality: > 0 means positive, < 0 negative. With ≥ or ≤ the zeros count as well, but points where the expression is undefined never do.
x \le -9 \;\text{ or }\; x \ge -8The zeros count too (≤ / ≥), but values where it is undefined never do.
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6 Solutionx \le -9 \;\text{ or }\; x \ge -8\qquad \left( -\infty, -9 \right] \cup \left[ -8, \infty \right)
Check with test values
| x | Left side | Right side | Holds? |
|---|---|---|---|
| −10 | 2 | 0 | ✓ true |
| −7 | 2 | 0 | ✓ true |
| −17/2 | −1/4 | 0 | ✗ false |
Change a number or try your own problem: the solver works it out the same way, with a graph and hint mode.