# Solve x^2 – 5x + 4 > 0.

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Solve x2 – 5x + 4 > 0.

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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, ∞).