If $\alpha ,\beta $ are the zeroes of the polynomial $f\left( x \right)={{x}^{2}}+x+1$ then $\dfrac{1}{\alpha }+\dfrac{1}{\beta }=\_\_\_\_\_\_\_\_$ .
(a) 1
(b) 0
(c) -1
(d) 2
Answer
638.7k+ views
Hint: For solving this question we will simply find the sum and the product of the given roots by comparing the given polynomial with $f\left( x \right)={{x}^{2}}-\left( \text{sum of the roots} \right)x+\left( \text{product of the roots} \right)$ to get the correct values of $\alpha +\beta $ and $\alpha \beta $ . After that we will find the value of $\dfrac{1}{\alpha }+\dfrac{1}{\beta }$ by dividing the value of $\alpha +\beta $ by $\alpha \beta $.
Complete step by step answer:
Given:
It is given that $\alpha ,\beta $ are the zeroes of the polynomial $f\left( x \right)={{x}^{2}}+x+1$ and we have to find the value of $\dfrac{1}{\alpha }+\dfrac{1}{\beta }$ .
Now, before we proceed we should know that if $x=\alpha $ and $x=\beta $ are roots of a quadratic polynomial $f\left( x \right)={{x}^{2}}-\left( \alpha +\beta \right)x+\alpha \beta $ and we can write it as $f\left( x \right)={{x}^{2}}-\left( \text{sum of the roots} \right)x+\left( \text{product of the roots} \right)$ .
Now, if $\alpha ,\beta $ are the zeroes of the polynomial $f\left( x \right)={{x}^{2}}+x+1$ then with the help of the above concept we conclude the following results:
$\begin{align}
& \alpha +\beta =-1.............\left( 1 \right) \\
& \alpha \beta =1....................\left( 2 \right) \\
\end{align}$
Now, just divide the equation (1) by equation (2). Then,
$\begin{align}
& \dfrac{\alpha +\beta }{\alpha \beta }=-1 \\
& \Rightarrow \dfrac{1}{\beta }+\dfrac{1}{\alpha }=-1 \\
& \Rightarrow \dfrac{1}{\alpha }+\dfrac{1}{\beta }=-1 \\
\end{align}$
Now, from the above result we conclude that if $\alpha ,\beta $ are the zeroes of the polynomial $f\left( x \right)={{x}^{2}}+x+1$ and we have to find the value of $\dfrac{1}{\alpha }+\dfrac{1}{\beta }$ will be equal to -1.
Hence, option (c) will be the correct option.
Note: Here, the student should first understand what is asked in the question and then proceed in the right direction to get the correct answer quickly. Moreover, though the problem is very easy, we should write sum and the product of the roots without any mistake and in the end, we should use our intelligence to find the correct answer quickly.
Complete step by step answer:
Given:
It is given that $\alpha ,\beta $ are the zeroes of the polynomial $f\left( x \right)={{x}^{2}}+x+1$ and we have to find the value of $\dfrac{1}{\alpha }+\dfrac{1}{\beta }$ .
Now, before we proceed we should know that if $x=\alpha $ and $x=\beta $ are roots of a quadratic polynomial $f\left( x \right)={{x}^{2}}-\left( \alpha +\beta \right)x+\alpha \beta $ and we can write it as $f\left( x \right)={{x}^{2}}-\left( \text{sum of the roots} \right)x+\left( \text{product of the roots} \right)$ .
Now, if $\alpha ,\beta $ are the zeroes of the polynomial $f\left( x \right)={{x}^{2}}+x+1$ then with the help of the above concept we conclude the following results:
$\begin{align}
& \alpha +\beta =-1.............\left( 1 \right) \\
& \alpha \beta =1....................\left( 2 \right) \\
\end{align}$
Now, just divide the equation (1) by equation (2). Then,
$\begin{align}
& \dfrac{\alpha +\beta }{\alpha \beta }=-1 \\
& \Rightarrow \dfrac{1}{\beta }+\dfrac{1}{\alpha }=-1 \\
& \Rightarrow \dfrac{1}{\alpha }+\dfrac{1}{\beta }=-1 \\
\end{align}$
Now, from the above result we conclude that if $\alpha ,\beta $ are the zeroes of the polynomial $f\left( x \right)={{x}^{2}}+x+1$ and we have to find the value of $\dfrac{1}{\alpha }+\dfrac{1}{\beta }$ will be equal to -1.
Hence, option (c) will be the correct option.
Note: Here, the student should first understand what is asked in the question and then proceed in the right direction to get the correct answer quickly. Moreover, though the problem is very easy, we should write sum and the product of the roots without any mistake and in the end, we should use our intelligence to find the correct answer quickly.
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