x2 – 5x + 4 > 0 ⇒ (x – 1) (x – 4) > 0
Now equating (x – 1) and (x – 4) to zero, we get the critical points as 1 and 4. Plot the points 1 and 4 on the real number line and then examine the sign of the expression in the three portions of the line divided by these points.
When \(x\) < 1, i.e., x ∈ (– ∞, 1), both the terms of the given expression are negative, hence (\(x\) – 1) (\(x\) – 4) > 0.
Similarly when \(x\) > 4, i.e., \(x\)∈ (4, ∞), (\(x\) – 1) (\(x\) – 4) > 0.
When \(x\)∈ (1, 4), one term being +ve and other –ve, the expression (\(x\) – 1) (\(x\) – 4) < 0.
∴ For (\(x\) – 1) (\(x\) – 4) > 0, the required solution set is \(x\)∈ (–∞, 1) ∪ (4, ∞).
