
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
160.8k+ views
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.
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.
Recently Updated Pages
Geometry of Complex Numbers – Topics, Reception, Audience and Related Readings

JEE Main 2021 July 25 Shift 1 Question Paper with Answer Key

JEE Main 2021 July 22 Shift 2 Question Paper with Answer Key

JEE Electricity and Magnetism Important Concepts and Tips for Exam Preparation

JEE Energetics Important Concepts and Tips for Exam Preparation

JEE Isolation, Preparation and Properties of Non-metals Important Concepts and Tips for Exam Preparation

Trending doubts
JEE Main 2025 Session 2: Application Form (Out), Exam Dates (Released), Eligibility, & More

JEE Main 2025: Derivation of Equation of Trajectory in Physics

Electric Field Due to Uniformly Charged Ring for JEE Main 2025 - Formula and Derivation

Displacement-Time Graph and Velocity-Time Graph for JEE

JEE Main 2026 Syllabus PDF - Download Paper 1 and 2 Syllabus by NTA

JEE Main Eligibility Criteria 2025

Other Pages
JEE Advanced Marks vs Ranks 2025: Understanding Category-wise Qualifying Marks and Previous Year Cut-offs

JEE Advanced 2025: Dates, Registration, Syllabus, Eligibility Criteria and More

JEE Advanced Weightage 2025 Chapter-Wise for Physics, Maths and Chemistry

Degree of Dissociation and Its Formula With Solved Example for JEE

Free Radical Substitution Mechanism of Alkanes for JEE Main 2025

JEE Advanced 2025 Notes
