Question

If the roots of the quadratic equation \[\begin{array}{*{20}{c}}
  {\dfrac{{x - m}}{{mx + 1}}}& = &{\dfrac{{x + n}}{{nx + 1}}}
\end{array}\] are reciprocal to each other, then
A) \[\begin{array}{*{20}{c}}
  n& = &0
\end{array}\]
B) \[\begin{array}{*{20}{c}}
  m& = &n
\end{array}\]
C) \[m + n\]
D) \[m - n\]

Answer 1

Hint: In this question, first of all, we will convert the given equation in the general form of the quadratic equation. After converting the equation, we will determine the product of the roots of the equation. Hence, we will get a suitable answer.



Formula Used:Formula used:
1) \[\begin{array}{*{20}{c}}
  {\alpha \beta }& = &{\dfrac{c}{a}}
\end{array}\]


Complete step by step solution:In this question, we have given a quadratic equation. Let us assume that the roots of the quadratic equation are \[\alpha \]and \[\beta \]respectively.
According to the condition that is given in the question, the roots of the quadratic equations are reciprocal to each other. If the one root of a quadratic equation is \[\alpha \], then the other root of the quadratic equation will be \[\dfrac{1}{\alpha }\]. Therefore, we can write the other root as,
\[\begin{array}{*{20}{c}}
  \beta & = &{\dfrac{1}{\alpha }}
\end{array}\]
Now we have the equation
\[ \Rightarrow \begin{array}{*{20}{c}}
  {\dfrac{{x - m}}{{mx + 1}}}& = &{\dfrac{{x + n}}{{nx + 1}}}
\end{array}\]
Now we will convert the above equation into the general form of the quadratic equation. For that purpose, first of all, we will do a cross multiplication of the above equation. Therefore, we will get
\[ \Rightarrow \begin{array}{*{20}{c}}
  {\left( {x - m} \right)\left( {nx + 1} \right)}& = &{\left( {mx + 1} \right)\left( {x + n} \right)}
\end{array}\]
\[ \Rightarrow \begin{array}{*{20}{c}}
  {n{x^2} + x - mnx - m}& = &{m{x^2} + mnx + x + n}
\end{array}\]
Now we will get
\[ \Rightarrow \begin{array}{*{20}{c}}
  {\left( {m - n} \right){x^2} + 2mnx + \left( {m + n} \right)}& = &0
\end{array}\]
Now we will determine the product of the roots of the above equation. Therefore, we know that
\[ \Rightarrow \begin{array}{*{20}{c}}
  {\alpha \beta }& = &{\dfrac{c}{a}}
\end{array}\]
Therefore,
\[ \Rightarrow \begin{array}{*{20}{c}}
  {\alpha \times \dfrac{1}{\alpha }}& = &{\dfrac{{m + n}}{{m - n}}}
\end{array}\]
\[ \Rightarrow \begin{array}{*{20}{c}}
  1& = &{\dfrac{{m + n}}{{m - n}}}
\end{array}\]
Now we will do the cross multiplication. Therefore, we will get
\[ \Rightarrow \begin{array}{*{20}{c}}
  {m + n}& = &{m - n}
\end{array}\]
Finally, we will get
\[\begin{array}{*{20}{c}}
  { \Rightarrow n}& = &0
\end{array}\]
Hence, we can choose the correct answer from the given options.



Option ‘A’ is correct

Note: In this question, it is important to note that we will get the answer when we get the product of the roots of the quadratic equation.