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(a) If $\alpha, \beta$ are zeroes of the polynomial $8x^2 - 5x - 1$, then form a quadratic polynomial in $x$ whose zeroes are $\frac{2}{\alpha}$ and $\frac{2}{\beta}$.
OR
(b) Find the zeroes of the polynomial $p(x) = 3x^2 + x - 10$ and verify the relationship between zeroes and its coefficients.
OR
(b) Find the zeroes of the polynomial $p(x) = 3x^2 + x - 10$ and verify the relationship between zeroes and its coefficients.
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(a) $p(x) = 8x^2 - 5x - 1$
$\alpha + \beta = \frac{5}{8}, \alpha\beta = \frac{-1}{8}$ ($\frac{1}{2} + \frac{1}{2}$ marks)
$\therefore \text{sum of zeroes} = \frac{2}{\alpha} + \frac{2}{\beta} = -10$ ($\frac{1}{2}$ mark)
$\text{and product of zeroes} = \frac{2}{\alpha} \times \frac{2}{\beta} = -32$ ($\frac{1}{2}$ mark)
$\text{Required polynomial} = x^2 + 10x - 32$ (1 mark)
OR
(b) $p(x) = 3x^2 + x - 10 = (x + 2)(3x - 5)$
$\text{Zeroes of } p(x) \text{ are } -2 \text{ and } \frac{5}{3}$ (1 mark)
$\text{Sum of zeroes} = -2 + \frac{5}{3} = \frac{-1}{3} = -\frac{\text{coefficient of } x}{\text{coefficient of } x^2}$ (1 mark)
$\text{Product of zeroes} = -2 \times \frac{5}{3} = \frac{-10}{3} = \frac{\text{constant term}}{\text{coefficient of } x^2}$ (1 mark)
$\alpha + \beta = \frac{5}{8}, \alpha\beta = \frac{-1}{8}$ ($\frac{1}{2} + \frac{1}{2}$ marks)
$\therefore \text{sum of zeroes} = \frac{2}{\alpha} + \frac{2}{\beta} = -10$ ($\frac{1}{2}$ mark)
$\text{and product of zeroes} = \frac{2}{\alpha} \times \frac{2}{\beta} = -32$ ($\frac{1}{2}$ mark)
$\text{Required polynomial} = x^2 + 10x - 32$ (1 mark)
OR
(b) $p(x) = 3x^2 + x - 10 = (x + 2)(3x - 5)$
$\text{Zeroes of } p(x) \text{ are } -2 \text{ and } \frac{5}{3}$ (1 mark)
$\text{Sum of zeroes} = -2 + \frac{5}{3} = \frac{-1}{3} = -\frac{\text{coefficient of } x}{\text{coefficient of } x^2}$ (1 mark)
$\text{Product of zeroes} = -2 \times \frac{5}{3} = \frac{-10}{3} = \frac{\text{constant term}}{\text{coefficient of } x^2}$ (1 mark)