Solve x² − 10x + 9 ≥ 0
Quadratic inequality, worked out line by line the way a teacher would write it.
Answer
| Solution | x ≤ 1 or x ≥ 9 |
| Interval notation | (−∞, 1] ∪ [9, ∞) |
Step-by-step solution
6 steps-
1 Givenx^{2} - 10 x + 9 \ge 0
Solve for x.
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2 Factor\left(x - 9\right) \left(x - 1\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 = 9,\; x = 1The 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,\, 1\right) & 1 & \left(1,\, 9\right) & 9 & \left(9,\, \infty\right) \\ \hline x - 9 & - & - & - & 0 & + \\ \hline x - 1 & - & 0 & + & + & + \\ \hline \text{whole} & + & 0 & - & 0 & + \end{array}Test values: x = 0, x = 2, x = 10.
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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 1 \;\text{ or }\; x \ge 9The zeros count too (≤ / ≥), but values where it is undefined never do.
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6 Solutionx \le 1 \;\text{ or }\; x \ge 9\qquad \left( -\infty, 1 \right] \cup \left[ 9, \infty \right)
Check with test values
| x | Left side | Right side | Holds? |
|---|---|---|---|
| 0 | 9 | 0 | ✓ true |
| 10 | 9 | 0 | ✓ true |
| 2 | −7 | 0 | ✗ false |
Change a number or try your own problem: the solver works it out the same way, with a graph and hint mode.