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(a) $\alpha, \beta$ are zeroes of the polynomial $3x^2 - 8x + k$. Find the value of $k$, if $\alpha^2 + \beta^2 = \frac{40}{9}$.
OR
(b) Find the zeroes of the polynomial $2x^2 + 7x + 5$ and verify the relationship between its zeroes and co-efficients.
OR
(b) Find the zeroes of the polynomial $2x^2 + 7x + 5$ and verify the relationship between its zeroes and co-efficients.
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(a) $p(x) = 3x^2 - 8x + k \Rightarrow \alpha + \beta = \frac{8}{3}, \alpha\beta = \frac{k}{3}$
$\alpha^2 + \beta^2 = \frac{40}{9} \Rightarrow (\frac{8}{3})^2 - \frac{2k}{3} = \frac{40}{9} \Rightarrow k = 4$
OR
(b) $p(x) = 2x^2 + 7x + 5 = (x + 1)(2x + 5)$
Zeroes of $p(x)$ are $-1$ and $-\frac{5}{2}$
Sum of zeroes $= -1 - \frac{5}{2} = -\frac{7}{2} = -\frac{\text{coefficient of } x}{\text{coefficient of } x^2}$
Product of zeroes $= (-1)(-\frac{5}{2}) = \frac{5}{2} = \frac{\text{constant term}}{\text{coefficient of } x^2}$
$\alpha^2 + \beta^2 = \frac{40}{9} \Rightarrow (\frac{8}{3})^2 - \frac{2k}{3} = \frac{40}{9} \Rightarrow k = 4$
OR
(b) $p(x) = 2x^2 + 7x + 5 = (x + 1)(2x + 5)$
Zeroes of $p(x)$ are $-1$ and $-\frac{5}{2}$
Sum of zeroes $= -1 - \frac{5}{2} = -\frac{7}{2} = -\frac{\text{coefficient of } x}{\text{coefficient of } x^2}$
Product of zeroes $= (-1)(-\frac{5}{2}) = \frac{5}{2} = \frac{\text{constant term}}{\text{coefficient of } x^2}$