Find the zeroes of the quadratic polynomial 7y^2 – 11y/3 – 2/3
Find the zeroes of the quadratic polynomial 7y2 – 11y/3 – 2/3 and verify the relationship between the zeroes and the coefficients.
Answer
p(y) = 7y2 – 11y/3 – 2/3
= 1/3 × (21y2 – 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