If $\alpha$ and $\beta$ are the zeroes of the quadratic polynomial $f(x) = x^2 - (\sqrt{5} -1)x -(\sqrt{5} + 1)$, then the value of $\displaystyle \frac{1}{\alpha^2} + \frac{1}{\beta^2}$ is ____________.
Sum of roots alpha + beta = sqrt(5) - 1. Product alpha * beta = -(sqrt(5) + 1). 1/alpha^2 + 1/beta^2 = (alpha^2 + beta^2) / (alpha * beta)^2 = ((alpha + beta)^2 - 2 * alpha * beta) / (alpha * beta)^2. Plugging in values: ((sqrt(5)-1)^2 + 2(sqrt(5)+1)) / (-(sqrt(5)+1))^2 = (5 + 1 - 2*sqrt(5) + 2*sqrt(5) + 2) / (5 + 1 + 2*sqrt(5)) = 8 / (6 + 2*sqrt(5)) = 4 / (3 + sqrt(5)). Rationalizing gives 4(3 - sqrt(5)) / (9 - 5) = 3 - sqrt(5).
Using the relationship between zeros and coefficients for f(x) = x^2 - (sqrt(5) - 1)x - (sqrt(5) + 1), the sum of the roots (alpha + beta) is sqrt(5) - 1 and the product (alpha * beta) is -(sqrt(5) + 1). We rewrite the required expression as (1/alpha^2) + (1/beta^2) = (alpha^2 + beta^2) / (alpha*beta)^2, which equals [(alpha + beta)^2 - 2(alpha*beta)] / (alpha*beta)^2. Substituting the values gives [(sqrt(5) - 1)^2 - 2(-sqrt(5) - 1)] / (-sqrt(5) - 1)^2, expanding to [(6 - 2*sqrt(5)) + 2*sqrt(5) + 2] / (6 + 2*sqrt(5)), which simplifies to 8 / (6 + 2*sqrt(5)). Rationalizing the denominator by multiplying by (6 - 2*sqrt(5)) results in 8(6 - 2*sqrt(5)) / 16, giving the final value of 3 - sqrt(5).