Problem 29 - 71 JEE Main Mathematics Online (2020 - 2012) & Offline (2018 - 2002) Chapterwise + Topicwise Solved Papers

Question

Let \(\alpha\) and \(\beta\) be the roots of equation \(\mathrm{px}^{2}+\mathrm{qx}+\mathrm{r}=0, \mathrm{p} \neq 0\). If \(\mathrm{p}, \mathrm{q}, \mathrm{r}\) are in A.P and \(\frac{1}{\alpha}+\frac{1}{\beta}=4\), then the value of \(|\alpha-\beta|\) is: (a) \(\frac{\sqrt{34}}{9}\) (b) \(\frac{2 \sqrt{13}}{9}\) (c) \(\frac{\sqrt{61}}{9}\) (d) \(\frac{2 \sqrt{17}}{9}\)

Step-by-Step Solution

Answer

The value of \(|\alpha - \beta|\) is \(\frac{2\sqrt{13}}{9}\).

Step 1: Understanding the Problem

We are given a quadratic equation \( px^2 + qx + r = 0 \) with roots \( \alpha \) and \( \beta \), and it's given that \( \frac{1}{\alpha} + \frac{1}{\beta} = 4 \). Additionally, \( p, q, r \) are in arithmetic progression (A.P.), meaning \( 2q = p + r \). We need to find the value of \( |\alpha - \beta| \).

Step 2: Using Vieta's Formulas

By Vieta's formulas for the roots of a quadratic equation, we have \( \alpha + \beta = -\frac{q}{p} \) and \( \alpha \beta = \frac{r}{p} \). We also know that \( \frac{1}{\alpha} + \frac{1}{\beta} = \frac{\alpha + \beta}{\alpha \beta} = 4 \).

Step 3: Substitute Vieta's Formulas into Given Condition

Using \( \frac{\alpha + \beta}{\alpha \beta} = 4 \), we substitute: \(-\frac{q}{p} \bigg/ \frac{r}{p} = 4\) which simplifies to \(-\frac{q}{r} = 4\). Therefore, \(q = -4r\).

Step 4: Identify the Relation Using A.P. Condition

Since \( p, q, r \) are in A.P., we have the equation \( 2q = p + r \). Substituting \( q = -4r \) into this, \( 2(-4r) = p + r \), which simplifies to \(-8r = p + r\), so \( p = -9r \).

Step 5: Finding \(|\alpha - \beta|\)

Recall that \( |\alpha - \beta| = \sqrt{(\alpha + \beta)^2 - 4\alpha \beta} \). Substitute \( \alpha + \beta = -\frac{q}{p} = \frac{4r}{9r} = \frac{4}{9} \) and \( \alpha \beta = \frac{r}{p} = \frac{r}{-9r} = -\frac{1}{9} \). Calculate: \[ |\alpha - \beta| = \sqrt{\left(\frac{4}{9}\right)^2 - 4\left(-\frac{1}{9}\right)} = \sqrt{\frac{16}{81} + \frac{4}{9}} \].

Step 6: Simplify to Find Absolute Difference

Convert \( \frac{4}{9} \) to have common denominator 81: \( \frac{4}{9} = \frac{36}{81} \). Therefore, \( |\alpha - \beta| = \sqrt{\frac{16}{81} + \frac{36}{81}} = \sqrt{\frac{52}{81}} = \frac{\sqrt{52}}{9} \). This reduces to \( \frac{2\sqrt{13}}{9} \).

Key Concepts

Vieta's Formulas

Vieta's formulas provide a fascinating connection between the coefficients of a quadratic equation and its roots. In a quadratic equation of the form \( ax^2 + bx + c = 0 \), the formulas tell us that the sum of the roots \( \alpha + \beta = -\frac{b}{a} \) and the product of the roots \( \alpha \beta = \frac{c}{a} \). These relationships are invaluable when solving problems involving quadratic equations.
In the exercise, Vieta's formulas help to express the given condition \( \frac{1}{\alpha} + \frac{1}{\beta} = 4 \) in terms of the equation's coefficients. This condition simplifies to \( \frac{\alpha + \beta}{\alpha \beta} = 4 \).
This substitution is possible because from Vieta's, \( \alpha + \beta = -\frac{q}{p} \) and \( \alpha \beta = \frac{r}{p} \). By manipulating these expressions, students can link algebraic insights to problem-solving techniques in quadratic equations.

Arithmetic Progression

An arithmetic progression (A.P.) is a sequence of numbers in which the difference between consecutive terms is constant, known as the common difference. For the sequence \( a_1, a_2, a_3, \ldots \), the formula for the \( n \)-th term is given by \( a_n = a_1 + (n-1)d \), where \( d \) is the common difference.
In the context of the exercise, the coefficients \( p, q, r \) of the quadratic equation are in A.P., meaning that the relationship \( 2q = p + r \) holds. This arises because the common difference between these terms is \((q - p) = (r - q)\).
Identifying and working with arithmetic progression in such problems helps simplify complex algebraic expressions and understand the relationships between coefficients in polynomial equations. Through A.P., this problem effectively links a sequence property to a condition involving the equation's structure.

Roots of a Quadratic Equation

The roots of a quadratic equation \( ax^2 + bx + c = 0 \) are the values of \( x \) which satisfy the equation. They can be found using the quadratic formula: \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \). These roots are essential in various calculations involving quadratic equations.
For the given exercise, understanding the roots is key to finding the absolute difference \( |\alpha - \beta| \). This difference is calculated as \( \sqrt{(\alpha + \beta)^2 - 4\alpha \beta} \).
By substituting Vieta's relationships derived from the coefficients, it becomes possible to compute \( |\alpha - \beta| \). The process involves manipulating the sum \( \alpha + \beta \) and product \( \alpha \beta \) derived from Vieta's formulas. Thus, mastery of manipulating and understanding roots within a quadratic's structure can significantly aid in solving algebraic problems.