iv) Given that, sum of zeroes (S)=−32√5
and product of zeroes (P)=−12
∴ Required quadratic expression,
F(x)=x2–Sx+P=x2+32√5x−12
=2√5x2+3x−√5
Using factorization method = 2√5x2+5x–2x–√5
=√5x(2x+√5)−1(2x+√5)
=(2x+√5)(√5x−1)
Hence, the zeroes of f(x) are −√52 and 1√5