141
Find zeroes of the polynomial $6x^2 - 7x - 3$ and verify the relationship between zeroes and its coefficients.
Show SolutionHide Solution↓
$p(x) = 6x^2 - 7x - 3 = (2x - 3)(3x + 1)$
Zeroes of $p(x)$ are $\frac{3}{2}$ and $-\frac{1}{3}$
Sum of zeroes $= \frac{3}{2} - \frac{1}{3} = \frac{7}{6} = -\frac{\text{coefficient of } x}{\text{coefficient of } x^2}$
Product of zeroes $= \frac{3}{2} \times \frac{-1}{3} = \frac{-3}{6} = \frac{\text{constant term}}{\text{coefficient of } x^2}$
Zeroes of $p(x)$ are $\frac{3}{2}$ and $-\frac{1}{3}$
Sum of zeroes $= \frac{3}{2} - \frac{1}{3} = \frac{7}{6} = -\frac{\text{coefficient of } x}{\text{coefficient of } x^2}$
Product of zeroes $= \frac{3}{2} \times \frac{-1}{3} = \frac{-3}{6} = \frac{\text{constant term}}{\text{coefficient of } x^2}$