Consider the system of equations y = x^2 + 4x + k and y = -2x + 5, where k...

1. TRANSLATE the problem setup

• Given information:
  • Two equations: \(\mathrm{y = x^2 + 4x + k}\) and \(\mathrm{y = -2x + 5}\)
  • \(\mathrm{k}\) is a positive integer
  • The system has no real solutions
• What "no real solutions" means: The graphs don't intersect anywhere

2. INFER the solution strategy

• To find intersections, we set the y-values equal to each other
• If there are no intersections, the resulting equation will have no real solutions
• This happens when a quadratic's discriminant is negative

3. SIMPLIFY by setting equations equal

Set \(\mathrm{x^2 + 4x + k = -2x + 5}\)

Move everything to one side:

\(\mathrm{x^2 + 4x + k + 2x - 5 = 0}\)

\(\mathrm{x^2 + 6x + (k - 5) = 0}\)

4. TRANSLATE the "no real solutions" condition

• For quadratic \(\mathrm{ax^2 + bx + c = 0}\), no real solutions means \(\mathrm{D \lt 0}\)
• Here: \(\mathrm{a = 1}\), \(\mathrm{b = 6}\), \(\mathrm{c = (k - 5)}\)
• So we need: \(\mathrm{6^2 - 4(1)(k - 5) \lt 0}\)

5. SIMPLIFY the discriminant inequality

\(\mathrm{36 - 4(k - 5) \lt 0}\)

\(\mathrm{36 - 4k + 20 \lt 0}\)

\(\mathrm{56 - 4k \lt 0}\)

\(\mathrm{4k \gt 56}\)

\(\mathrm{k \gt 14}\)

6. APPLY CONSTRAINTS to find the answer

• \(\mathrm{k}\) must be a positive integer
• \(\mathrm{k \gt 14}\) means \(\mathrm{k ≥ 15}\)
• The least possible value is \(\mathrm{k = 15}\)

Answer: 15

Why Students Usually Falter on This Problem

Most Common Error Path:

Weak TRANSLATE skill: Students don't connect "no real solutions" to the discriminant condition. They might try to solve the system directly or graph the functions instead of recognizing that "no solutions" means \(\mathrm{D \lt 0}\). This leads to confusion and guessing.

Second Most Common Error:

Poor SIMPLIFY execution: Students make algebraic errors when combining like terms (getting \(\mathrm{x^2 + 2x}\) instead of \(\mathrm{x^2 + 6x}\)) or when manipulating the discriminant inequality. A common mistake is getting the inequality direction wrong, leading to \(\mathrm{k \lt 14}\) instead of \(\mathrm{k \gt 14}\). This causes them to select an answer that's too small.

The Bottom Line:

This problem requires connecting the geometric concept of "no intersections" to the algebraic concept of "negative discriminant." Students who miss this connection often struggle to even begin the problem systematically.