Question

Sum of roots is \[ - 1\] and sum of their reciprocals is \[\frac{1}{6}\] ​ , then equation is?
A) \[\begin{array}{*{20}{c}}
  {{x^2} + x - 6}& = &0
\end{array}\]
B) \[\begin{array}{*{20}{c}}
  {{x^2} - x + 6}& = &0
\end{array}\]
C) \[\begin{array}{*{20}{c}}
  {6{x^2} + x + 1}& = &0
\end{array}\]
D) \[\begin{array}{*{20}{c}}
  {{x^2} - 6x + 1}& = &0
\end{array}\]

Answer 1

Hint: In this question, we have given the sum of the roots of an equation and the sum of their reciprocals. Therefore, first of all, we will determine the product of the roots of an unknown equation. After that, we will use the general form of the quadratic equation when the sum and the product of the roots are given. Hence, we will get a suitable answer.

Formula used:
If the roots of equation are \[\alpha \] and \[\beta \] then equation is:
 \[\begin{array}{*{20}{c}}
  {{x^2} - \left( {\alpha + \beta } \right)x + \alpha \beta }& = &0
\end{array}\]

Complete step by step solution:
Let us assume that there is an unknown equation whose roots are \[\alpha \]and \[\beta \]. And we have given the sum of the roots. Therefore, we can write
\[\begin{array}{*{20}{c}}
  { \Rightarrow \alpha + \beta }& = &{ - 1}
\end{array}\] --------- (1)
And we have also given the sum of their reciprocal. Therefore, we can write
\[ \Rightarrow \begin{array}{*{20}{c}}
  {\frac{1}{\alpha } + \frac{1}{\beta }}& = &{\frac{1}{6}}
\end{array}\]
Therefore,
\[ \Rightarrow \begin{array}{*{20}{c}}
  {\frac{{\beta + \alpha }}{{\alpha \beta }}}& = &{\frac{1}{6}}
\end{array}\]
Now put the value of equation (1) in the above equation. So, we will get
\[ \Rightarrow \begin{array}{*{20}{c}}
  {\frac{{ - 1}}{{\alpha \beta }}}& = &{\frac{1}{6}}
\end{array}\]
\[ \Rightarrow \begin{array}{*{20}{c}}
  {\alpha \beta }& = &{ - 6}
\end{array}\] -------- (2).
Now we know the general form of the quadratic equation when the sum and the product of the roots are given. Therefore,
\[ \Rightarrow \begin{array}{*{20}{c}}
  {{x^2} - \left( {\alpha + \beta } \right)x + \alpha \beta }& = &0
\end{array}\]
Now put the value of equations of (1) and (2) in the above equation. Therefore, we will get
\[ \Rightarrow \begin{array}{*{20}{c}}
  {{x^2} - \left( { - 1} \right)x - 6}& = &0
\end{array}\]
\[ \Rightarrow \begin{array}{*{20}{c}}
  {{x^2} + x - 6}& = &0
\end{array}\]
Now the final answer is \[\begin{array}{*{20}{c}}
  {{x^2} + x - 6}& = &0
\end{array}\].

Therefore, the correct option is A.

Note: In this question, the first point is to keep in mind that whenever we have to find the quadratic equation if the roots are given, then always determine the sum and the product of the roots of the equation.