Question

# If $\alpha ,\beta$ are the zeroes of the polynomials f(x) = ${x^2} + x + 1$ then $\dfrac{1}{\alpha } + \dfrac{1}{\beta }$ is ………………A. 1B. 0C. -1D. 2

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Hint - To solve such types of questions we need to know the basic formulae of sum of roots and product of roots of the quadratic equation.

Given,${x^2} + x + 1$
On comparing with standard quadratic equation, $a{x^{{2^{}}}} + bx + c = 0$
Sum of roots ($\alpha + \beta$ ) =$\dfrac{{ - b}}{a}$ and Products of roots ($\alpha .\beta$ ) =$\dfrac{c}{a}$
Sum of roots ($\alpha + \beta$ )= $\dfrac{{ - 1}}{1} = - 1$ and Product of roots ($\alpha .\beta$ )= $\dfrac{1}{1} = 1$ ……..(i)
We need to find the value of $\dfrac{1}{\alpha } + \dfrac{1}{\beta }$
On solving, taking L.C.M. we get $\dfrac{{\alpha + \beta }}{{\alpha .\beta }}$
=$\dfrac{{ - 1}}{1} = - 1$ $\dfrac{{ - 1}}{1} = - 1$
Note - For solving such type of quadratic question we need to find the value of sum and product of its roots and then solved the given relation and putting the value of sum and product of roots to solve the equation $a{x^{{2^{}}}} + bx + c = 0$ then sum of roots ($\alpha + \beta$ ) =$\dfrac{{ - b}}{a}$ and Products of roots ($\alpha .\beta$ ) =$\dfrac{c}{a}$.