Introduction:
A quadratic polynomial is an expression of the form ax² + bx + c, where a, b, and c are constants. The zeroes of a quadratic polynomial are the values of x that make the polynomial equal to zero. In this problem, we are given a quadratic polynomial and we need to find the value of k such that the zeroes of the polynomial are negative reciprocals of each other.

Method:
To find the zeroes of the given quadratic polynomial, we can use the quadratic formula. The quadratic formula states that the zeroes of a quadratic polynomial ax² + bx + c are given by:

x = (-b ± √(b² - 4ac)) / 2a

If the zeroes of the polynomial are negative reciprocals of each other, then we can write:

x₁ = -1/x₂

Substituting this into the quadratic formula, we get:

-1/x₂ = (-b ± √(b² - 4ac)) / 2a

Multiplying both sides by x₂, we get:

-1 = (-b ± √(b² - 4ac))x₂ / 2a

Solving for x₂, we get:

x₂ = (-2a) / (-b ± √(b² - 4ac))

Since we want the zeroes of the polynomial to be negative reciprocals of each other, we can set x₁ = -1/x₂ and solve for k.

Solution:
The given quadratic polynomial is:

p(x) = 2x² + 9x (k - 2k + 1)

Expanding the polynomial, we get:

p(x) = 2x² + 9x (k - 2k + 1)
= 2x² + 9x (1 - k)

The discriminant of the quadratic formula is:

b² - 4ac = (9(1-k))² - 4(2)(0)

Simplifying, we get:

b² - 4ac = 81(1 - 2k)

Since we want the zeroes of the polynomial to be negative reciprocals of each other, we can set x₁ = -1/x₂ and solve for k. Substituting the values for the zeroes from the quadratic formula, we get:

x₁ = (-9 ± √(81(1 - 2k))) / 4

x₂ = (-2) / (-9 ± √(81(1 - 2k)))

Setting x₁ = -1/x₂ and simplifying, we get:

(9 ± √(81(1 - 2k))) / 2 = -2 / x₂

Multiplying both sides by x₂ and simplifying, we get:

(9 ± √(81(1 - 2k)))x₂ = -4

Substituting the value of x₂ from the quadratic formula, we get:

(9 ± √(81(1 - 2k))) (-9 ± √(81(1 - 2k))) = -8(1 - k)

Simplifying, we get:

81(1 - 2k) - 81(1 - 2k) ± 18√(81(1 - 2k)) =