147
Find the zeroes of the polynomial $p(x) = 4x^2 - 8x + 3$ and verify the relationship between its zeroes and co-efficients.
Show SolutionHide Solution↓
Solution:
(a) $p(x) = 4x^2 - 8x + 3$
Zeroes of $p(x)$ are $\frac{3}{2}, \frac{1}{2}$ (1 Mark)
Sum of zeroes = $\frac{3}{2} + \frac{1}{2} = 2 = -(\frac{-8}{4}) = -\frac{\text{coefficient of } x}{\text{coefficient of } x^2}$ (1 Mark)
Product of zeroes = $\frac{3}{2} \times \frac{1}{2} = \frac{3}{4} = \frac{\text{coefficient of } x^0}{\text{coefficient of } x^2}$ (1 Mark)
(a) $p(x) = 4x^2 - 8x + 3$
Zeroes of $p(x)$ are $\frac{3}{2}, \frac{1}{2}$ (1 Mark)
Sum of zeroes = $\frac{3}{2} + \frac{1}{2} = 2 = -(\frac{-8}{4}) = -\frac{\text{coefficient of } x}{\text{coefficient of } x^2}$ (1 Mark)
Product of zeroes = $\frac{3}{2} \times \frac{1}{2} = \frac{3}{4} = \frac{\text{coefficient of } x^0}{\text{coefficient of } x^2}$ (1 Mark)