Question

If $\alpha \text{ and }\beta $ are roots of the polynomial $f\left( x \right)=5{{y}^{2}}-7y+1,$ find the value of $\dfrac{1}{\alpha }+\dfrac{1}{\beta }$.

Answer 1

Hint: For solving this problem, we obtain the sum and product of roots individually from the given equation and then find the ratio of the sum of zeroes and the product of zeroes, by this we can easily find the value of $\dfrac{1}{\alpha }+\dfrac{1}{\beta }$ as on simplification it is equivalent to the ratio of sum and product of roots.

Complete Step-by-Step solution:
In algebra, a quadratic function is a polynomial function with one or more variables in which the highest-degree term is of the second degree. A single-variable quadratic function can be stated as:
$f(x)=a{{x}^{2}}+bx+c,\quad a\ne 0$
If we have two zeroes of a quadratic equation then the polynomial could be formed by using the simplified result which could be stated as:
${{x}^{2}}-(\alpha +\beta )x+\alpha \beta $, where $\alpha \text{ and }\beta $ are two zeroes of the equation.
In the question f(x) can be stated as: $f\left( x \right)=5{{y}^{2}}-7y+1$
Divide the equation by 5, to obtain equation similar to one stated above,
$\Rightarrow f\left( x \right)={{y}^{2}}-\dfrac{7}{5}y+\dfrac{1}{5}$
Compare the above equation from ${{x}^{2}}-(\alpha +\beta )x+\alpha \beta $, we get
$\begin{align}
  & \therefore \alpha +\beta =\dfrac{7}{5}...\left( 1 \right) \\
 & \therefore \alpha \beta =\dfrac{1}{5}...\left( 2 \right) \\
\end{align}$
Dividing the equation (1) and equation (2), we get
$\begin{align}
  & \Rightarrow \dfrac{\alpha +\beta }{\alpha \beta }=\dfrac{\dfrac{7}{5}}{\dfrac{1}{5}} \\
 & \Rightarrow \dfrac{\alpha +\beta }{\alpha \beta }=\dfrac{7\times 5}{5} \\
 & \Rightarrow \dfrac{\alpha +\beta }{\alpha \beta }=\dfrac{7}{1}...\left( 3 \right) \\
\end{align}$
So, the ratio of sum and product of roots is 7.
Now the required expression can be simplified as, $\Rightarrow \dfrac{1}{\alpha }+\dfrac{1}{\beta }=\dfrac{\alpha +\beta }{\alpha \beta }$
The value of $\dfrac{\alpha +\beta }{\alpha \beta }$ from equation (3) is:
$\dfrac{\alpha +\beta }{\alpha \beta }=7$
Hence, the value of $\dfrac{1}{\alpha }+\dfrac{1}{\beta }$ is 7.

Note: This problem can be alternatively solved by obtaining the value of roots by using the discriminant formula which can be stated as: $Roots=\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}$. By using this expression, we have the value of roots and we can easily evaluate $\dfrac{1}{\alpha }+\dfrac{1}{\beta }$.