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Find the zeroes of the polynomial $25a^2 - 10a + 1$ and verify the relationship between the zeroes and coefficients of the given polynomial.
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Solution: $25a^2 - 10a + 1 = (5a - 1) (5a - 1)$
Zeroes are $\frac{1}{5}$ and $\frac{1}{5}$
Sum of zeroes $= \frac{1}{5} + \frac{1}{5} = \frac{2}{5} = \frac{-(-10)}{25} = \frac{-\text{Coefficient of } a}{\text{Coefficient of } a^2}$
Product of zeroes $= \frac{1}{5} \times \frac{1}{5} = \frac{1}{25} = \frac{\text{Constant term}}{\text{Coefficient of } a^2}$
Zeroes are $\frac{1}{5}$ and $\frac{1}{5}$
Sum of zeroes $= \frac{1}{5} + \frac{1}{5} = \frac{2}{5} = \frac{-(-10)}{25} = \frac{-\text{Coefficient of } a}{\text{Coefficient of } a^2}$
Product of zeroes $= \frac{1}{5} \times \frac{1}{5} = \frac{1}{25} = \frac{\text{Constant term}}{\text{Coefficient of } a^2}$