Solve the inequality x2 + 2x > 15.

Analyze the question: We need all values of x for which x² + 2x > 15 (reading "x2" as x²).

Concept: Move everything to one side, factorise, then use the sign of the quadratic. Since the x² coefficient is positive, the parabola opens upward, so the expression is positive outside the two roots.

Solution:

1. Rearrange: x² + 2x − 15 > 0

2. Factorise (find two numbers that multiply to −15 and add to +2: that is +5 and −3): (x + 5)(x − 3) > 0

3. Critical values: x = −5 and x = 3

4. Test the three regions: • x < −5 (try x = −6): (−1)(−9) = 9 > 0 ✓ • −5 < x < 3 (try x = 0): (5)(−3) = −15 < 0 ✗ • x > 3 (try x = 4): (9)(1) = 9 > 0 ✓

5. Solution: x < −5 or x > 3, which can be written as x > 3 or x < −5.

Checking the options:

• A. x < −3 or x > 5: wrong roots (it comes from x² − 2x − 15)
• B. −5 < x < 3: this is where the expression is negative
• C. x < 3 or x > 5: wrong
• D. x > 3 or x < −5: matches our result ✓

Tip: For "> 0" with a positive x² coefficient, the answer is always "outside the roots" (x < smaller or x > larger). For "< 0", it is "between the roots". Always sanity-check with one test value, such as x = 0 here: 0 > 15 is false, so 0 must not be in the solution, which rules out B.