Consider the quadratic inequality x^2 - 7x + 12 > 0. By analysing the corresponding quadratic equation and its factorisation, which of the following ranges of x correctly represents the full solution set of this inequality?

Introduction / Context:
Quadratic inequalities are a standard topic in aptitude and algebra. The typical approach is to first solve the corresponding quadratic equation to find critical points (roots), then use the sign of the quadratic on different intervals. This question asks you to interpret x^2 - 7x + 12 > 0 and choose the correct range of x that satisfies the inequality, testing your understanding of factorisation and sign analysis.

Given Data / Assumptions:

• The inequality is x^2 - 7x + 12 > 0.
• We consider real values of x.
• We must choose the correct description of the solution set from the options.

Concept / Approach:
First, solve the related equation x^2 - 7x + 12 = 0 by factorisation. The roots split the number line into intervals. Because the coefficient of x^2 is positive, the parabola opens upward. Therefore, the quadratic expression will be positive outside the interval between the roots and negative inside it. Once we know the roots and understand the shape of the graph, we can write the solution set of x^2 - 7x + 12 > 0 as the union of appropriate intervals.

Step-by-Step Solution:
Consider the equation x^2 - 7x + 12 = 0. Factorise the quadratic: x^2 - 7x + 12 = (x - 3)(x - 4). Set each factor equal to zero to find roots: x - 3 = 0 or x - 4 = 0. So the roots are x = 3 and x = 4. Because the coefficient of x^2 is positive, the parabola opens upwards. The expression (x - 3)(x - 4) is zero at x = 3 and x = 4, negative between these roots, and positive outside this interval. We want x^2 - 7x + 12 > 0, that is, (x - 3)(x - 4) > 0. The product of two factors is positive when both are positive or both are negative. Both negative when x < 3. Both positive when x > 4. Therefore, the solution set is x < 3 or x > 4.

Verification / Alternative check:
Pick test points in each region. For x = 2 (less than 3), x^2 - 7x + 12 = 4 - 14 + 12 = 2, which is positive. For x = 3.5 (between 3 and 4), we get 3.5^2 - 7 * 3.5 + 12 = 12.25 - 24.5 + 12 = -0.25, which is negative. For x = 5 (greater than 4), x^2 - 7x + 12 = 25 - 35 + 12 = 2, which is positive again. This matches our interval reasoning.

Why Other Options Are Wrong:
3 < x < 4 corresponds to the region where the quadratic is negative, so it does not satisfy the inequality with greater than zero. Saying x < 3 only or x > 4 only ignores half of the valid solution set. The option 3 ≤ x ≤ 4 describes the region where the expression is zero or negative, not strictly positive.

Common Pitfalls:
Students often confuse the direction of the inequality and pick the region between the roots instead of outside them. Another mistake is to forget that strict inequality (>) excludes the roots themselves, although in this question the options are written in interval form that already reflects strict inequalities. Always analyse the sign on each interval carefully.

Final Answer:
The solution set of the inequality is x < 3 or x > 4.