74
Find the zeroes of the polynomial $2t^2 - 9t - 45$ and verify the relationship between the zeroes and the coefficients of the polynomial.
Show SolutionHide Solution↓
Sol. $2t^2 - 9t - 45 = 2t^2 - 15t + 6t - 45$
$= (2t - 15) (t + 3)$
$\therefore$ zeroes of the polynomial are $\frac{15}{2}$ and $-3$.
Sum of the zeroes = $\frac{15}{2} + (-3) = \frac{9}{2} = -\frac{\text{coefficient of } t}{\text{coefficient of } t^2}$
Product of the zeroes = $\frac{15}{2} \times (-3) = -\frac{45}{2} = \frac{\text{constant term}}{\text{coefficient of } t^2}$
$= (2t - 15) (t + 3)$
$\therefore$ zeroes of the polynomial are $\frac{15}{2}$ and $-3$.
Sum of the zeroes = $\frac{15}{2} + (-3) = \frac{9}{2} = -\frac{\text{coefficient of } t}{\text{coefficient of } t^2}$
Product of the zeroes = $\frac{15}{2} \times (-3) = -\frac{45}{2} = \frac{\text{constant term}}{\text{coefficient of } t^2}$