For what value of k, the zeroes of x2 + kx + 12 will differ by 1?

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

a=24; b=12

step-by-step explanation:

a/b=2: 1

a=2b

36/(1+2)=12

b=12

a=24

answer: y = -2.5x - 15

              (-6,0)

step-by-step explanation: