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Hint: Zeros of the quadratic equation are the values of the dependent variable for which the quadratic expression becomes 0.

Complete step-by-step answer:

To find the value of zeros, put f(x)=0.

$ \Rightarrow $ ${x^2} - 5x + 4 = 0$

$ \Rightarrow $ ${x^2} - 4x - x + 4 = 0$

$ \Rightarrow $ $x(x - 4) -1 (x - 4) = 0$

\[ \Rightarrow \left( {x - 1} \right)\left( {x - 4} \right) = 0\]

Zeros of the quadratic polynomial are

$ \Rightarrow \alpha = 1, \beta = 4$

Now,

$ \Rightarrow \dfrac{1}{\alpha } + \dfrac{1}{\beta } - 2\alpha \beta = \dfrac{1}{1} + \dfrac{1}{4} - 2 \times 1 \times 4$

$ \Rightarrow \dfrac{5}{4} - 8$

$ \Rightarrow - \dfrac{{27}}{4}$

Note: Zeros is the intersection of the polynomial and the axis, if the polynomial is in x, then zeros is the intersection of the polynomial with x-axis. The roots can also be found using the quadratic formula.

Complete step-by-step answer:

To find the value of zeros, put f(x)=0.

$ \Rightarrow $ ${x^2} - 5x + 4 = 0$

$ \Rightarrow $ ${x^2} - 4x - x + 4 = 0$

$ \Rightarrow $ $x(x - 4) -1 (x - 4) = 0$

\[ \Rightarrow \left( {x - 1} \right)\left( {x - 4} \right) = 0\]

Zeros of the quadratic polynomial are

$ \Rightarrow \alpha = 1, \beta = 4$

Now,

$ \Rightarrow \dfrac{1}{\alpha } + \dfrac{1}{\beta } - 2\alpha \beta = \dfrac{1}{1} + \dfrac{1}{4} - 2 \times 1 \times 4$

$ \Rightarrow \dfrac{5}{4} - 8$

$ \Rightarrow - \dfrac{{27}}{4}$

Note: Zeros is the intersection of the polynomial and the axis, if the polynomial is in x, then zeros is the intersection of the polynomial with x-axis. The roots can also be found using the quadratic formula.

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