Question

# If $\alpha$ and $\beta$ are the zeros of the quadratic polynomial $f\left( x \right) = {x^2} - 5x + 4$, find the value of $\dfrac{1}{\alpha } + \dfrac{1}{\beta } - 2\alpha \beta$.

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

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.