How to Solve Quadratic Inequalities
Quadratic inequalities ask you to find the range of values of x that make a quadratic expression positive or negative, rather than equal to zero. On this page you will learn a clear, reliable method for how to solve quadratic inequalities: rearrange so the quadratic is compared with zero, solve to find the critical values, sketch the parabola, and read the solution set straight off the graph. We cover the single-range case (the answer is between the roots), the two-range case (the answer is outside the roots), and what changes when the coefficient of x squared is negative. Worked examples show each step, and the practice room gives unlimited randomised questions across five difficulty levels, from finding the roots to mixed positive and negative coefficients, all auto-marked instantly.
What is a quadratic inequality?
A quadratic inequality looks like \(x^2 - x - 6 < 0\) or \(x^2 - 5x + 4 \ge 0\): a quadratic expression compared with \(0\) using \(<\), \(>\), \(\le\) or \(\ge\). The answer is not a single value but a range (or two ranges) of \(x\).
The reliable method is to think about the graph. Write \(y\) for the quadratic expression, so \(y = ax^2 + bx + c\). Solving the inequality means finding where \(y\) is negative or positive.
Here "the axis" always means the x-axis, the horizontal line where \(y = 0\). When the curve is below the x-axis, \(y\) is negative (\(y < 0\)); when it is above the x-axis, \(y\) is positive (\(y > 0\)). Every quadratic with \(a > 0\) is a U-shaped parabola, so it dips below the x-axis between its roots and rises above it outside the roots.
The four-step method
- Make one side zero if it is not already (rearrange so the quadratic is compared with \(0\)).
- Find the critical values by solving the equation \(ax^2 + bx + c = 0\). These are where the parabola crosses the x-axis.
- Sketch the parabola (U-shaped for \(a > 0\)) and mark the roots.
- Read off the region: if you want \(y < 0\) (curve below the x-axis) the answer is one range between the roots; if you want \(y > 0\) (curve above the x-axis) the answer is two ranges outside the roots.
The two outcomes
Why the graph tells you the answer
Take \(y = x^2 - x - 6 = (x-3)(x+2)\), with roots at \(x = -2\) and \(x = 3\). The parabola is below the x-axis between the roots and above it outside them.
Less than 0: below the x-axis
Solve \(x^2-x-6 < 0\): want \(y<0\), so the region below the x-axis, between the roots. Answer: \(-2 < x < 3\)
Greater than 0: above the x-axis
Solve \(x^2-x-6 > 0\): want \(y>0\), so the region above the x-axis, outside the roots. Answer: \(x < -2\) or \(x > 3\)
When \(a\) is negative: the sad parabola flips it
Take \(y = -x^2 + x + 6 = -(x-3)(x+2)\), same roots \(-2\) and \(3\) but opening downward. Now the curve is above the x-axis between the roots and below it outside, the opposite way round. This is why you should always sketch and read the actual graph rather than memorise a rule.
Greater than 0: above the x-axis
Solve \(-x^2+x+6 > 0\): want \(y>0\), so above the x-axis, which is now between the roots. Answer: \(-2 < x < 3\)
Less than 0: below the x-axis
Solve \(-x^2+x+6 < 0\): want \(y<0\), so below the x-axis, which is now outside the roots. Answer: \(x < -2\) or \(x > 3\)
Worked examples
💡 Example 1: less than zero (one range)
Solve \(x^2 - 2x - 8 < 0\).
Solve the equation to find the critical values:
\[x^2 - 2x - 8 = 0\]
\[(x-4)(x+2) = 0\]
\(x = 4\) or \(x = -2\)
The parabola is U-shaped, so it is below the axis between the roots.
Solution: \(-2 < x < 4\)
What's happening?
The inequality is \(< 0\), meaning we want where the curve dips below the x-axis.
That is the single region between the two roots, so the answer is one inequality.
Strict \(<\), so the roots are not included.
💡 Example 2: greater than or equal (two ranges)
Solve \(x^2 - 3x - 18 \ge 0\).
Solve the equation:
\[x^2 - 3x - 18 = 0\]
\[(x-6)(x+3) = 0\]
\(x = 6\) or \(x = -3\)
U-shaped, so it is above the axis outside the roots.
Solution: \(x \le -3\) or \(x \ge 6\)
What's happening?
The inequality is \(\ge 0\), meaning we want where the curve is on or above the x-axis.
That is the two outer regions, so the answer is two inequalities joined by "or".
Inclusive \(\ge\), so the roots are included.
💡 Example 3: rearrange first
Solve \(x^2 < 49\).
Make one side zero:
\[x^2 - 49 < 0\]
\[(x-7)(x+7) = 0\]
\(x = 7\) or \(x = -7\)
\(< 0\) means below the axis, between the roots.
Solution: \(-7 < x < 7\)
What's happening?
It is tempting to write \(x < 7\), but that misses the negative values.
Rearranging to compare with \(0\) first, then using the graph, gives the full range, including down to \(-7\).
🔑 Key points
- Always compare with \(0\) before finding the critical values.
- \(< 0\) or \(\le 0\): one range, between the roots.
- \(> 0\) or \(\ge 0\): two ranges, outside the roots.
- Sketch the U-shaped parabola to decide which region you want.
- Closed circles and the roots included only for \(\le\) and \(\ge\).
⚠ Common pitfalls
- Square-rooting \(x^2 < 49\) to get only \(x < 7\), losing the negative end.
- Writing the final answer as an equation (\(=\)) instead of keeping the inequality.
- Joining two separate ranges into one impossible inequality like \(3 < x < 2\).
- Assuming \(> 0\) always means "outside the roots": when the coefficient of \(x^2\) is negative the parabola opens downward, so above and below the x-axis swap over. Sketch it every time.