Question

If roots of \[{x^2} - 7x + 6 = 0\] are \[\alpha ,\beta \], then \[\dfrac{1}{\alpha } + \dfrac{1}{\beta }\]
A) $\dfrac{6}{7}$
B) $\dfrac{7}{6}$
C) $\dfrac{7}{{10}}$
D) $\dfrac{8}{9}$

Answer 1

Hint: In this question, we are given \[\alpha ,\beta \] are the roots of the quadratic equation \[{x^2} - 7x + 6 = 0\] and we have to find the value of the sum of the reciprocal of the given roots. Firstly, using the formula of the sum and the product of the roots calculate the sum and product. Then, take the L.C.M. of \[\dfrac{1}{\alpha } + \dfrac{1}{\beta }\] and put the required values. Solve it further.

Formula Used:
General quadratic equation: – \[A{x^2} + Bx + C = 0\]
Let, the roots of the above quadratic equation be $\alpha $ and $\beta $
Therefore,
Sum of roots, $\alpha + \beta = \dfrac{{ - B}}{A}$
Product of roots, $\alpha \beta = \dfrac{C}{A}$

Complete step by step Solution:
Given that,
The roots of the quadratic equation \[{x^2} - 7x + 6 = 0\] are $\alpha $ and $\beta $.
Compare the above equation \[{x^2} - 7x + 6 = 0\] with the general quadratic equation \[A{x^2} + Bx + C = 0\]
We get,
$A = 1$, $B = - 7$, and $C = 6$
Now, using the formula of sum and product of the roots i.e., Sum of roots, $\alpha + \beta = \dfrac{{ - B}}{A}$ and product of roots, $\alpha \beta = \dfrac{C}{A}$
It implies that,
$\alpha + \beta = 7$ and $\alpha \beta = 6$ --------(1)
Now, we’ll solve the given expression \[\dfrac{1}{\alpha } + \dfrac{1}{\beta }\]
Here, the L.C.M. of $\alpha $ and $\beta $ is $\alpha \beta $
Therefore, \[\dfrac{1}{\alpha } + \dfrac{1}{\beta } = \dfrac{{\alpha + \beta }}{{\alpha \beta }}\]
Now, using equation (1) substitute the required values.
It will be \[\dfrac{1}{\alpha } + \dfrac{1}{\beta } = \dfrac{7}{6}\]
Thus, the value of \[\dfrac{1}{\alpha } + \dfrac{1}{\beta }\] is equal to \[\dfrac{7}{6}\].

Hence, the correct option is (B).

Note: The values of $x$ that fulfil a given quadratic equation \[A{x^2} + Bx + C = 0\] are known as its roots. They are, in other words, the values of the variable $\left( x \right)$ that satisfy the equation. The roots of a quadratic function are the $x - $ coordinates of the function's $x - $ intercepts. Because the degree of a quadratic equation is $2$, it can only have two roots.