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(a) If $\alpha, \beta$ are zeroes of the polynomial $3x^2 - 8x + 4$, then form a quadratic polynomial in $x$ whose zeroes are $\frac{1}{\alpha}$ and $\frac{1}{\beta}$.
OR
(b) Find zeroes of the polynomial $6x^2 - 7x - 3$ and verify the relationship between zeroes and its coefficients.
OR
(b) Find zeroes of the polynomial $6x^2 - 7x - 3$ and verify the relationship between zeroes and its coefficients.
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Solution: (a) $p(x) = 3x^2 - 8x + 4$
$\alpha + \beta = \frac{8}{3}, \alpha\beta = \frac{4}{3}$
$\therefore \frac{1}{\alpha} + \frac{1}{\beta} = \frac{\alpha + \beta}{\alpha\beta} = 2$ and $\frac{1}{\alpha\beta} = \frac{3}{4}$
$\therefore$ required polynomial is $x^2 - 2x + \frac{3}{4}$ or $k(4x^2 - 8x + 3)$, where $k$ is a non-zero real number.
OR
(b) $p(x) = 6x^2 - 7x - 3 = (2x - 3)(3x + 1)$
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 $= \frac{3}{2} \times \frac{-1}{3} = \frac{-3}{6} = \frac{\text{constant term}}{\text{coefficient of } x^2}$
$\alpha + \beta = \frac{8}{3}, \alpha\beta = \frac{4}{3}$
$\therefore \frac{1}{\alpha} + \frac{1}{\beta} = \frac{\alpha + \beta}{\alpha\beta} = 2$ and $\frac{1}{\alpha\beta} = \frac{3}{4}$
$\therefore$ required polynomial is $x^2 - 2x + \frac{3}{4}$ or $k(4x^2 - 8x + 3)$, where $k$ is a non-zero real number.
OR
(b) $p(x) = 6x^2 - 7x - 3 = (2x - 3)(3x + 1)$
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 $= \frac{3}{2} \times \frac{-1}{3} = \frac{-3}{6} = \frac{\text{constant term}}{\text{coefficient of } x^2}$