A quadratic inequality such as \(x^{2} - x - 12 < 0\) asks where a parabola lies below the \(x\)-axis. Find the critical values where it crosses, sketch the curve, then read off the part you need and write it in the correct form: one interval such as \(-3 < x < 4\), or two pieces such as \(x < 1\) or \(x > 6\).
To solve a quadratic inequality, rearrange it to \(f(x) > 0\), \(f(x) < 0\), \(f(x) \ge 0\) or \(f(x) \le 0\), solve \(f(x) = 0\) to get the critical values, and sketch \(y = f(x)\). For a U-shaped curve with critical values \(\alpha < \beta\): \(f(x) < 0\) gives ONE interval \(\alpha < x < \beta\) (between the roots), and \(f(x) > 0\) gives TWO pieces \(x < \alpha\) or \(x > \beta\) (outside the roots). Use \(\le\) and \(\ge\) when the inequality includes equality.
Fig. 1: y = x² - x - 12 = (x + 3)(x - 4) crosses the x-axis at the critical values -3 and 4. The curve is above the axis (y > 0) for x < -3 or x > 4, and below it (y < 0) for -3 < x < 4. Open circles mark the critical values, which are not included in a strict inequality
Section 01
Critical values, then sketch.
In the previous lesson you solved \(x^{2} - x - 12 = 0\) and found \(x = -3\) or \(x = 4\). Those two roots are the critical values of the inequality \(x^{2} - x - 12 < 0\): they are the only places where the sign of \(x^{2} - x - 12\) can change. A quick sketch tells you which side of each critical value you want.
Step 1. Make sure one side is zero, for example \(x^{2} - x - 12 < 0\).
Step 2. Solve the equation \(x^{2} - x - 12 = 0\) to get the critical values \(-3\) and \(4\).
Step 3. Sketch \(y = x^{2} - x - 12\): a U shape (positive \(x^{2}\) term) crossing the \(x\)-axis at \(-3\) and \(4\). You do not need the vertex.
Step 4. \(< 0\) means below the axis, \(> 0\) means above it. Read off the \(x\)-values of that part of the curve (Fig. 1).
A strict inequality (\(<\) or \(>\)) does not include the critical values; a non-strict one (\(\le\) or \(\ge\)) does, because there \(f(x) = 0\) and \(0 \le 0\) is true. On a number line show this with an open circle or a filled dot.
Worked example Between the roots
Solve \(x^{2} - x - 12 < 0\).
Critical values: \(x^{2} - x - 12 = (x + 3)(x - 4) = 0\), so \(x = -3\) or \(x = 4\).
The graph is a U shape crossing the \(x\)-axis at \(-3\) and \(4\).
\(< 0\) means the part of the curve below the axis, which lies between the critical values.
Check with \(x = 0\): \(0 - 0 - 12 = -12 < 0\), true, and \(0\) is between \(-3\) and \(4\).
Answer: \(-3 < x < 4\)
Worked example Outside the roots, with equality
Find the set of values of \(x\) for which \(2x^{2} + 5x - 3 \ge 0\).
An answer outside the roots is TWO separate inequalities joined by 'or'. Never write it as \(4 < x < -3\) (no number is both more than 4 and less than \(-3\)) or as 'x > 4 and x < -3'. An answer between the roots is one combined inequality, \(-3 < x < 4\).
Section 02
Make one side zero first.
The sketch only works when you compare the quadratic with zero. So collect every term on one side before you look for critical values, and keep the \(x^{2}\) term positive if you can: then the curve is always a U shape and 'below the axis' always means 'between the roots'.
Negative \(x^{2}\) term. Either sketch an upside-down (n-shaped) curve, or multiply both sides by \(-1\) and REVERSE the inequality sign: \(6 + x - x^{2} > 0\) becomes \(x^{2} - x - 6 < 0\).
Never divide by \(x\). \(x\) could be negative (which would reverse the sign) or zero, so dividing loses solutions. Bring everything to one side and factorise instead.
Worked example Expand, collect, factorise
Solve \(x(x + 2) > 15\).
Expand and make one side zero: \(x^{2} + 2x - 15 > 0\).
Critical values: \((x + 5)(x - 3) = 0\), so \(x = -5\) or \(x = 3\).
U shape; \(> 0\) means above the axis, outside the critical values.
Check with \(x = 0\): \(0 > 15\) is false, and \(0\) is correctly excluded.
Answer: \(x < -5\) or \(x > 3\)
Worked example A negative x-squared term
Solve \(6 + x - x^{2} > 0\).
Multiply both sides by \(-1\) and reverse the sign: \(x^{2} - x - 6 < 0\).
Critical values: \((x - 3)(x + 2) = 0\), so \(x = 3\) or \(x = -2\).
U shape; \(< 0\) means between the critical values.
Graph check: \(y = 6 + x - x^{2}\) is n-shaped and is above the axis between \(-2\) and \(3\), the same answer.
Answer: \(-2 < x < 3\)
Your turn
Solve \(x^{2} \ge 4x\).
\(x^{2} - 4x \ge 0\), so \(x(x - 4) \ge 0\). Critical values \(0\) and \(4\); outside them, including the ends: \(x \le 0\) or \(x \ge 4\). (Dividing by \(x\) would give only \(x \ge 4\) and lose \(x \le 0\).)
Examiner trap
Multiplying or dividing an inequality by a negative number reverses the sign. Students who turn \(6 + x - x^{2} > 0\) into \(x^{2} - x - 6 > 0\) get \(x < -2\) or \(x > 3\), exactly the wrong region. Test one value, such as \(x = 0\), in the ORIGINAL inequality to catch this.
Stage 1 · Learn
Check your understanding
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Solve \((x - 2)(x + 5) < 0\).
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Which is the solution of \(x^{2} - 16 \ge 0\)?
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Solve \(x^{2} > 3x\).
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What is the solution set of \(x^{2} + 2x + 5 < 0\)?
Section 03
Squares, no roots and contexts.
Some inequalities look simpler than they are. \(x^{2} < 9\) has two critical values, not one, because both \(3\) and \(-3\) square to 9. And a quadratic with no real roots never crosses the \(x\)-axis, so it is either positive for every \(x\) or negative for every \(x\).
Worked example A square on one side
Solve (a) \(x^{2} < 9\) (b) \(x^{2} > 9\).
Make one side zero: \(x^{2} - 9 < 0\), so \((x + 3)(x - 3) < 0\). Critical values \(-3\) and \(3\).
(a) \(< 0\): between the critical values, \(-3 < x < 3\).
(b) \(> 0\): outside the critical values, \(x < -3\) or \(x > 3\).
Check (a) with \(x = -4\): \(16 < 9\) is false, so \(x < 3\) alone would be wrong.
Answer: (a) \(-3 < x < 3\) (b) \(x < -3\) or \(x > 3\)
Worked example No critical values
Show that \(x^{2} - 4x + 7 > 0\) for all real \(x\), and state the solution set of \(x^{2} - 4x + 7 < 0\).
In the next lesson a condition on a constant \(k\) will turn into \(k^{2} - 8k + 12 < 0\). Solve it.
\((k - 2)(k - 6) < 0\): between the critical values, so \(2 < k < 6\). It is the same skill with \(k\) in place of \(x\).
Examiner trap
Square rooting both sides of \(x^{2} < 9\) to get \(x < 3\) (or \(x < \pm 3\), which means nothing) loses the lower end. Always rearrange to \(x^{2} - 9 < 0\), factorise and sketch: the answer is \(-3 < x < 3\).
Stage 2 · Exam
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Find the set of values of \(x\) for which \(2x^{2} - 7x + 3 > 0\).
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Solve \(x(2x + 1) \le 6\).
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Solve \(12 - x - x^{2} > 0\).
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Find the integer values of \(x\) that satisfy \(x^{2} - 2x - 8 < 0\).
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