Linear & Quadratic Inequalities
Question
Find the set of values of \(x\) for which \(x^2 - 4x + 4 \lt 0\)
Solution
Show solution Hide solution Fully worked — 4 steps
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Factorise\[ x^2 - 4x + 4 \lt 0 \]\[ (x - 2)(x - 2) \lt 0 \]
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Case 1 — 1st factor is positive, 2nd is negative\[ x - 2 \gt 0 \quad \text{and} \quad x - 2 \lt 0 \]
Add 2 to both sides of the 1st and 2nd inequality:
\[ x - 2 + 2 \gt 0 + 2 \quad \text{and} \quad x - 2 + 2 \lt 0 + 2 \]Combine similar terms:
\[ x \gt 2 \quad \text{and} \quad x \lt 2 \]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 \(2\) with a ray extending right (\(x \gt 2\)). The two rays point in opposite directions from the same point and share no values, so the inequalities have no common solution.
Number line graph of case 1: rays in opposite directions from open circles at 2 — no overlap, so the solution set is ∅ Solution set of case 1: \(\varnothing\) or null set
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Case 2 — 1st factor is negative, 2nd is positive\[ x - 2 \lt 0 \quad \text{and} \quad x - 2 \gt 0 \]
Add 2 to both sides of the 1st and 2nd inequality:
\[ x - 2 + 2 \lt 0 + 2 \quad \text{and} \quad x - 2 + 2 \gt 0 + 2 \]Combine similar terms:
\[ x \lt 2 \quad \text{and} \quad x \gt 2 \]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 \(2\) with a ray extending left (\(x \lt 2\)). The two rays point in opposite directions from the same point and share no values, so the inequalities have no common solution.
Number line graph of case 2: rays in opposite directions from open circles at 2 — no overlap, so the solution set is ∅ Solution set of case 2: \(\varnothing\) or null set
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Solution set
\(\therefore\) the solution set of the inequality \(x^2 - 4x + 4 \lt 0\) is \(\varnothing\) or null set. That is, the solution set is the empty set: no value of \(x\) satisfies the inequality.