Question

Find the zeroes of the following quadratic polynomial and verify the relationship between the zeroes and the coefficients.

4u² + 8u

Answer 1

Given polynomial is 4u² + 8u 

We have, 4u² + 8u = 4u (u + 2) 

The value of 4u² + 8u is 0, 

when the value of 4u(u + 2) = 0, i.e., 

when u = 0 or u + 2 = 0, i.e., 

when u = 0 (or) u = – 2 

∴ The zeroes of 4u² + 8u are 0 and – 2. 

Therefore, sum of the zeroes = 0 + (-2) = -2

= \(-\frac{Coefficient\,of\,u}{Coefficient\,of\,u^2}=\frac{-8}{4}=-2\)

And product of the zeroes 0 . (-2) = 0

\(=\frac{Constant\,term}{Coefficient\,of u^2}=\frac{0}{4}={0}\)