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If $\alpha, \beta$ are zeroes of the polynomial $p(x) = 5x^2 - 7x - 3$, then form a quadratic polynomial whose zeroes are $\frac{2}{\alpha}$ and $\frac{2}{\beta}$.
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Solution: (a) $\alpha + \beta = -\frac{(-7)}{5} = \frac{7}{5}$, $\alpha\beta = \frac{-3}{5}$ (
frac{1}{2} +
frac{1}{2} Mark)
Sum of zeroes of required polynomial = $\frac{2}{\alpha} + \frac{2}{\beta} = \frac{2(\alpha + \beta)}{\alpha\beta} = \frac{2(\frac{7}{5})}{-\frac{3}{5}} = -\frac{14}{3}$ (
frac{1}{2} Mark)
Product of zeroes of the required polynomial = $\frac{2}{\alpha} \times \frac{2}{\beta} = \frac{4}{\alpha\beta} = \frac{4}{-\frac{3}{5}} = -\frac{20}{3}$ (
frac{1}{2} Mark)
Required polynomial is $k(x^2 - (\text{sum of zeroes})x + \text{product of zeroes})$
$x^2 - (-\frac{14}{3})x + (-\frac{20}{3}) = x^2 + \frac{14}{3}x - \frac{20}{3}$ or $k(3x^2 + 14x - 20)$ where $k$ is any non-zero real number. (1 Mark)
frac{1}{2} +
frac{1}{2} Mark)
Sum of zeroes of required polynomial = $\frac{2}{\alpha} + \frac{2}{\beta} = \frac{2(\alpha + \beta)}{\alpha\beta} = \frac{2(\frac{7}{5})}{-\frac{3}{5}} = -\frac{14}{3}$ (
frac{1}{2} Mark)
Product of zeroes of the required polynomial = $\frac{2}{\alpha} \times \frac{2}{\beta} = \frac{4}{\alpha\beta} = \frac{4}{-\frac{3}{5}} = -\frac{20}{3}$ (
frac{1}{2} Mark)
Required polynomial is $k(x^2 - (\text{sum of zeroes})x + \text{product of zeroes})$
$x^2 - (-\frac{14}{3})x + (-\frac{20}{3}) = x^2 + \frac{14}{3}x - \frac{20}{3}$ or $k(3x^2 + 14x - 20)$ where $k$ is any non-zero real number. (1 Mark)