109
Find the zeroes of the polynomial $q(x) = 6x^2 - 5x - 1$ and verify the relationship between the zeroes of $q(x)$ and its coefficients.
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Solution: $q(x) = 6x^2 - 5x - 1$
$(6x + 1) (x - 1) = 0$
Zeroes are $x = -\frac{1}{6}, 1$ [1 mark]
Sum of zeroes $= -\frac{1}{6} + 1 = \frac{5}{6} = \frac{- \text{Coefficient of } x}{\text{Coefficient of } x^2}$ [1 mark]
Product of zeroes $= -\frac{1}{6} \times 1 = -\frac{1}{6} = \frac{\text{Constant term}}{\text{Coefficient of } x^2}$ [1 mark]
$(6x + 1) (x - 1) = 0$
Zeroes are $x = -\frac{1}{6}, 1$ [1 mark]
Sum of zeroes $= -\frac{1}{6} + 1 = \frac{5}{6} = \frac{- \text{Coefficient of } x}{\text{Coefficient of } x^2}$ [1 mark]
Product of zeroes $= -\frac{1}{6} \times 1 = -\frac{1}{6} = \frac{\text{Constant term}}{\text{Coefficient of } x^2}$ [1 mark]