Problem 13 - Algebra 1 Course in Mathematics for the IIT-JEE and Other Engineering Exams

Question

If the roots of the quadratic equation \(\frac{x-m}{m x+1}\) \(=\frac{x+n}{n x+1}\) are reciprocal to each other, then [MPPET-2001] (a) \(n=0\) (b) \(m=n\) (c) \(m+n=1\) (d) \(m^{2}+n^{2}=1\)

Step-by-Step Solution

Answer

(a) \(n=0\)

Step 1: Clear the Denominators

To eliminate the fractions from the given equation, we cross-multiply: \((x-m)(nx+1) = (x+n)(mx+1)\).

Step 2: Expand and Simplify

Expand both sides: \(nx^2 - mx - mnx - m = mx^2 + nx + m + nx\). Simplify to get: \((n-m)x^2 - (mn+1)x - m = 0\).

Step 3: Set Condition for Reciprocal Roots

If the roots are reciprocal, then their product is 1. In a quadratic equation \(ax^2 + bx + c = 0\), the product of the roots \(r_1 \cdot r_2 = \frac{c}{a}\). Set \(\frac{-m}{n-m} = 1\).

Step 4: Solve the Equation

Solving \(-m = n-m\), this simplifies to \(-m + m = n\), which simplifies further to \(n = 0\).

Key Concepts

Reciprocal Roots

When the roots of a quadratic equation are reciprocal, it essentially means that if one root is a number, say \( r \), the other root will be \( \frac{1}{r} \). This relationship implies that their product is equal to 1. In any quadratic equation of the form \( ax^2 + bx + c = 0 \), the product of the roots is given by \( \frac{c}{a} \). Therefore, for the product of roots to be 1, the equation must satisfy this condition.
Understanding reciprocal roots is crucial because it sets a specific condition involving the coefficients of the quadratic equation. By knowing that the product must be 1, we can derive other properties or manipulate the equation to find specific solutions.

Cross-Multiplication Technique

Cross-multiplication is a handy algebraic tool especially when dealing with equations containing fractions. It eliminates the fractions by multiplying each side of the equation by the denominators of the other side.
This is useful because it converts a fractional equation into a polynomial one, thereby simplifying the complexity of the problem. For example, in the original exercise, the equation \(\frac{x-m}{mx+1}=\frac{x+n}{nx+1}\) is simplified by cross-multiplying to get a quadratic form, \((x-m)(nx+1) = (x+n)(mx+1)\). This technique is widely used to make equations more manageable and bring them into a form where standard algebraic methods can be applied.

Product of Roots

The product of the roots of a quadratic equation can shed valuable light on the nature of the equation. For a quadratic in the form \( ax^2 + bx + c = 0 \), the product of its roots \( r_1 \cdot r_2 \) is calculated as \( \frac{c}{a} \).
In the context of reciprocal roots, the condition set forth by this concept provides a means of validation or solution. If given that the roots are reciprocal, then by setting the product \( r_1 \cdot r_2 = 1 \), we can solve for unknowns in the equation. For example, with the equation \((n-m)x^2 - (mn+1)x - m = 0\), this concept is applied to determine \( n \).

Quadratic Formula Manipulation

Manipulating the quadratic formula can be a powerful approach when resolving quadratic equations, especially in complex cases. The quadratic formula \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \) allows for the precise finding of roots, but manipulation of equations often refers to adjusting or simplifying quadratic equations to fit specific conditions or reveal particular characteristics of the roots.
In some scenarios, it's not just about finding the roots but reformulating the equation until it satisfies certain standards or uncovers particular results, like reciprocal roots. The adjustment process is part of this manipulation, as we see in setting specific conditions derived from the properties of quadratic roots.