# If $\alpha$ and $\beta$ are zeroes of $x^2-4x+1$, then find the value of $\frac{1}{\alpha}+\frac{1}{\beta}-\alpha\beta$.

Given: $\alpha$ and $\beta$ are zeroes of $x^2-4x+1$.

To do: To find the value of $\frac{1}{\alpha}+\frac{1}{\beta}-\alpha\beta$.

Solution:

Given quadratic polynomial is $x^2-4x+1$.

$\alpha$ and $\beta$ are zeroes of given quadratic polynomial.

$\therefore$ Sum of the zeroes$=\alpha+\beta=-( \frac{-4}{1})=4$

Product of the zeroes$=\alpha\beta=\frac{1}{1}=1$

$\therefore \frac{1}{\alpha}+\frac{1}{\beta}-\alpha\beta=\frac{\alpha+\beta}{\alpha\beta}-\alpha\beta$

$=\frac{4}{1}-1$

$=3$

Thus, $\frac{1}{\alpha}+\frac{1}{\beta}-\alpha\beta=3$.

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