Solve x$^2$ + 3x - 4  ≤ 0

Analyze the question: We need the values of x for which x² + 3x − 4 ≤ 0.

Concept: For a quadratic with a positive x² coefficient, the graph is a U-shaped parabola. It is below or on the x-axis (≤ 0) between its two roots.

Solution:

1. Factorise x² + 3x − 4. We need two numbers that multiply to −4 and add to +3. These are +4 and −1.

(x + 4)(x − 1) ≤ 0

2. The roots are x = −4 and x = 1.

3. Test the regions: • x < −4 (try x = −5): (−1)(−6) = 6 > 0, so this fails. • −4 < x < 1 (try x = 0): (4)(−1) = −4 < 0, so this works. • x > 1 (try x = 2): (6)(1) = 6 > 0, so this fails.

4. Because the inequality is "≤", the endpoints are included:

−4 ≤ x ≤ 1

Checking the options:

• A. "−4 ≥ x ≥ 1" would require x ≤ −4 and x ≥ 1 at the same time, which is impossible.
• B. "4 ≤ x ≤ −1" is also impossible.
• C. "4 ≥ x ≥ 1" gives the wrong interval.
• D. −4 ≤ x ≤ 1 matches our solution.

Tip: For a quadratic with a positive x² coefficient, "≤ 0" means between the roots, and "≥ 0" means outside the roots. Also, a correctly written range always has the smaller number on the left (for example, −4 ≤ x ≤ 1). If an option has the smaller number on the wrong side, it is probably a trap.