Let a and b be the zeroes of x² + kx + 12 such that |a - b| = 1.
By the factor theorem, we can write the quadratic in terms of its zeroes as
x² + kx + 12 = (x - a) (x - b)
Expand the right side and equate the coefficients:
x² + kx + 12 = x² - (a + b) x + ab
Then
a + b = -k
ab = 12
The condition that |a - b| = 1 has two cases, so without loss of generality assume a > b, so that |a - b| = a - b.
Then if a - b = 1, we get b = a - 1. Substitute this into the equations above and solve for k :
a + (a - 1) = -k → 2a = 1 - k → a = (1 - k)/2
a (a - 1) = 12 → (1 - k)/2 • ((1 - k)/2 - 1) = 12
→ (1 - k)²/4 - (1 - k)/2 = 12
→ (1 - k)² - 2 (1 - k) = 48
→ (1 - 2k + k²) - 2 (1 - k) = 48
→ k² - 1 = 48
→ k² = 49
→ k = ± √(49) = ±7
