Find the zeroes of the quadratic polynomial 7y^{2} – 11y/3 – 2/3 and verify the relationship between the zeroes and the coefficients.

### Answer

p(y) = 7y^{2} – 11y/3 – 2/3

= 1/3 × (21y^{2} – 11y – 2)

= 1/3 × [(7y + 1)(3y – 2)]

∴ Zeroes are 2/3, –1/7

Sum of zeroes = 2/3 – 1/7 = 11/21

–b/a = 11/21

∴ sum of zeroes = –b/a

Product of zeroes = (2/3)(–1/7) = – 2/21

c/a = (–2/3)(1/7) = – 2/21

∴ Product = c/a