Question

Find the zeroes of the quadratic polynomial (5y² + 10y) and verify the relation between the zeroes and the coefficients.

Answer 1

We have, 

f (u) = 5u² + 10u 

It can be written as 5u (u+2) 

∴ f (u) = 0 ⇒ 5u = 0 or u + 2 = 0 

⇒ u = 0 or u = −2 

So, the zeroes of f (u) are −2 and 0.

Sum of the zeroes = −2 + 0 =−2 = \(\frac{-2\times5}{1\times5}\) = \(\frac{-10}5\) = \(\frac{-(coefficient\,of\,x)}{(coefficient\,of\,u^2)}\)

Product of zeroes = −2 × 0 = 0 = \(\frac{0\times5}{1\times5}\) = \(\frac{-0}5\) = \( \frac{constant\,term}{(coefficient\,of\,u^2)}\)