Linear & Quadratic Inequalities
Question
Find the set of values of \(x\) for which \(x^2 - 3x - 10 \gt 0\)
Solution
Show solution Hide solution Fully worked — 4 steps
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Factorise\[ x^2 - 3x - 10 \gt 0 \]\[ (x - 5)(x + 2) \gt 0 \]
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Case 1 — both factors are positive\[ x - 5 \gt 0 \quad \text{and} \quad x + 2 \gt 0 \]
Add 5 to both sides of the 1st inequality and subtract 2 from both sides of the 2nd inequality:
\[ x - 5 + 5 \gt 0 + 5 \quad \text{and} \quad x + 2 - 2 \gt 0 - 2 \]Combine similar terms:
\[ x \gt 5 \quad \text{and} \quad x \gt -2 \]Graph to see the solution set: on the number line, an open circle at \(-2\) with a ray extending right (\(x \gt -2\)) and an open circle at \(5\) with a ray extending right (\(x \gt 5\)). Both inequalities hold from \(5\) rightwards — the intersection, marked by the thick segment on the number line.
Number line graph of case 1: open circles at −2 and 5 with rays right; the overlap from 5 rightwards gives the solution set x > 5 Solution set of case 1: \(x \gt 5\)
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Case 2 — both factors are negative\[ x - 5 \lt 0 \quad \text{and} \quad x + 2 \lt 0 \]
Add 5 to both sides of the 1st inequality and subtract 2 from both sides of the 2nd inequality:
\[ x - 5 + 5 \lt 0 + 5 \quad \text{and} \quad x + 2 - 2 \lt 0 - 2 \]Combine similar terms:
\[ x \lt 5 \quad \text{and} \quad x \lt -2 \]Graph to see the solution set: on the number line, an open circle at \(-2\) with a ray extending left (\(x \lt -2\)) and an open circle at \(5\) with a ray extending left (\(x \lt 5\)). Both inequalities hold from \(-2\) leftwards — the intersection, marked by the thick segment on the number line.
Number line graph of case 2: open circles at −2 and 5 with rays left; the overlap from −2 leftwards gives the solution set x < −2 Solution set of case 2: \(x \lt -2\)
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Solution set
\(\therefore\) the solution set of the inequality \(x^2 - 3x - 10 \gt 0\) is \(x \lt -2\) or \(x \gt 5\). That is, the solution set is every value of \(x\) less than minus 2, together with every value of \(x\) greater than 5.