I know only the first solution,

Solution: Given polynomial = 9x² - 1

⇒9x² - 3x + 3x - 1

⇒3x(3x - 1) + 1(3x - 1)

⇒(3x - 1) (3x + 1)

To obtain the zeroes of the polynomial

⇒(3x - 1)(3x + 1) = 0

⇒3x - 1 = 0 or 3x + 1 = 0

⇒3x = 1 or 3x = -1

⇒x = 1/3 or x = -1/3

∴The zeroes are α = 1/3 and β= -1/3.

for the polynomial given to us, a = 9 , b = 0 and c= -1

Verifying the relation between the zeroes and the coefficients of the given polynomial.

1) α + β = -b/a

⇒1/3 + [-1/3] = -0/9

⇒1/3 - 1/3 =0

⇒0 = 0.

2) αβ = c/a

⇒1/3[-1/3 = -1/9

⇒-1/9 = -1/9

Hence verified.