If \[\alpha \], \[\beta \] are the roots of \[\begin{array}{*{20}{c}}
  {9{x^2} + 6x + 1}& = &0
\end{array}\], then the equation with the roots \[\dfrac{1}{\alpha }\], \[\dfrac{1}{\beta }\] is:
A) \[\begin{array}{*{20}{c}}
  {2{x^2} + 3x + 18}& = &0
\end{array}\]
B) \[\begin{array}{*{20}{c}}
  {{x^2} + 6x - 9}& = &0
\end{array}\]
C) \[\begin{array}{*{20}{c}}
  {{x^2} + 6x + 9}& = &0
\end{array}\]
D) \[\begin{array}{*{20}{c}}
  {{x^2} - 6x + 9}& = &0
\end{array}\]

Hint: In this question, first of all, we will determine the sum and the product of the roots of the equation. And we have given the roots of an unknown equation so we will also determine the sum and the product of the roots of an unknown equation. After that, we will apply the general form of the equation when the sum and the product of the roots of the equation are given. Hence, we'll get a suitable answer.

Formula Used:1) \[\begin{array}{*{20}{c}}
  {\alpha + \beta }& = &{ - \dfrac{b}{a}}
\end{array}\]
2) \[\begin{array}{*{20}{c}}
  {\alpha \beta }& = &{\dfrac{c}{a}}
\end{array}\]
3) \[{x^2} - \left( {\alpha + \beta } \right)x + \alpha \beta \]



Complete step by step solution:According to the question, we have given the equation whose roots are \[\alpha \]and \[\beta \]respectively. Therefore,
\[\begin{array}{*{20}{c}}
  { \Rightarrow 9{x^2} + 6x + 1}& = &0
\end{array}\]
Now the sum and the product of the above equation will be
\[\begin{array}{*{20}{c}}
  { \Rightarrow \alpha + \beta }& = &{ - \dfrac{6}{9}}
\end{array}\]
\[\begin{array}{*{20}{c}}
  { \Rightarrow \alpha + \beta }& = &{ - \dfrac{2}{3}}
\end{array}\] --------- (1)
And
\[\begin{array}{*{20}{c}}
  { \Rightarrow \alpha \beta }& = &{\dfrac{1}{9}}
\end{array}\] ----------- (2)
Now we have given an unknown equation whose roots are given. Therefore, the sum and the product of the roots of that unknown equation will be,
\[ \Rightarrow \dfrac{1}{\alpha } + \dfrac{1}{\beta }\]
\[ \Rightarrow \dfrac{{\alpha + \beta }}{{\alpha \beta }}\]
Now we will put the value of equations (1) and (2). Therefore, we will get
\[ \Rightarrow - \dfrac{2}{3} \times \dfrac{9}{1}\]
\[ \Rightarrow - 6\]
And
\[ \Rightarrow \dfrac{1}{\alpha } \times \dfrac{1}{\beta }\]
\[ \Rightarrow \dfrac{1}{{\alpha \beta }}\]
Now put the value of equation (1) in the above equation. Therefore, we will get
\[ \Rightarrow \dfrac{1}{{\dfrac{1}{9}}}\]
\[ \Rightarrow 9\]
Now we will put the value of the sum and the product of the roots of an unknown equation in the general form of the equation as
\[\begin{array}{*{20}{c}}
  { \Rightarrow {x^2} - \left( {\dfrac{1}{\alpha } + \dfrac{1}{\beta }} \right)x + \dfrac{1}{{\alpha \beta }}}& = &0
\end{array}\]
Therefore, we will get
\[\begin{array}{*{20}{c}}
  { \Rightarrow {x^2} + 6x + 9}& = &0
\end{array}\]



Option ‘C’ is correct

Note: In this question, the first point is to keep in mind that to determine the unknown equation, we will use the sum and the product of the roots of the unknown equation.