Let the zeroes of the polynomial x2 – x – 1 be α and β
We have,

Now, according to the given condition,
α2 β2 = ( – 1)2 = 1
& (α + β)2 = α2 + β2 + 2 α β
α2 + β2 = (α + β)2 – 2αβ
α2 + β2 = (1)2 – 2( – 1)
α2 + β2 = 3
So, the quadratic polynomial
= x2 – (sum of zeroes)x + product of zeroes
= x2 – (3)x + 1
= x2 – 3x + 1