Solve x² − x − 56 < 0
Quadratic inequality, worked out line by line the way a teacher would write it.
Answer
| Solution | −7 < x < 8 |
| Interval notation | (−7, 8) |
Step-by-step solution
6 steps-
1 Givenx^{2} - x - 56 < 0
Solve for x.
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2 Factor\left(x - 8\right) \left(x + 7\right) < 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 = -7The 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,\, -7\right) & -7 & \left(-7,\, 8\right) & 8 & \left(8,\, \infty\right) \\ \hline x - 8 & - & - & - & 0 & + \\ \hline x + 7 & - & 0 & + & + & + \\ \hline \text{whole} & + & 0 & - & 0 & + \end{array}Test values: x = -8, x = -6, x = 9.
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5 Keep the intervals where the expression is negative
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.
-7 < x < 8The critical points are left out: there the expression is 0 or undefined, not negative.
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6 Solution-7 < x < 8\qquad \left( -7, 8 \right)
Check with test values
| x | Left side | Right side | Holds? |
|---|---|---|---|
| −6 | −14 | 0 | ✓ true |
| −8 | 16 | 0 | ✗ false |
Change a number or try your own problem: the solver works it out the same way, with a graph and hint mode.