Problem 64 - A Complete Resource Book in Mathematics for JEE Main 2017

Question

If \(\alpha, \beta\) are the roots of the equation \(x^{2}+p x+q=0\) then \(\frac{\alpha}{\beta}\) is a root of the equation (A) \(p x^{2}+\left(2 q-p^{2}\right) x+p=0\) (B) \(q x^{2}+\left(p^{2}-2 q\right) x+q=0\) (C) \(q x^{2}+\left(2 q-p^{2}\right) x+q=0\) (D) None of these

Step-by-Step Solution

Answer

The root \( \frac{\alpha}{\beta} \) is a solution of equation (C).

Step 1: Understand the Problem

We need to find an equation for which \( \frac{\alpha}{\beta} \) is a root, given \( \alpha \) and \( \beta \) are the roots of the quadratic equation \( x^2 + px + q = 0 \).

Step 2: Apply Viète's Formulas

Using Viète's formulas, we know \( \alpha + \beta = -p \) and \( \alpha \beta = q \).

Step 3: Evaluate the Expression \( \frac{\alpha}{\beta} \)

Consider \( y = \frac{\alpha}{\beta} \), then \( \alpha = y \beta \). Substitute \( \alpha = y \beta \) and \( \alpha \beta = q \) into these equations.

Step 4: Substitute in Viète’s Formulas

Since \( \alpha + \beta = -p \), we substitute \( y \beta + \beta = -p \). This gives us \( (y + 1) \beta = -p \).

Step 5: Solve for \( \beta \)

Express \( \beta = \frac{-p}{y + 1} \) and use \( \alpha \beta = q \), giving us \( y \left(\frac{-p}{y + 1}\right)^2 = q \). This converts to \( q y + 2pq + q = 0 \).

Step 6: Compare with Given Equation Choices

Rearrange the terms to form a quadratic equation: \( qy^2 + (2q - p^2)y + q = 0 \), matching option (C).

Key Concepts

Viète's Formulas

Viète's Formulas provide a way to relate the coefficients of a quadratic equation to the sum and product of its roots. Let's consider a quadratic equation of the form \(x^2 + px + q = 0\). In this equation:

• The sum of the roots \(\alpha\) and \(\beta\) is \(\alpha + \beta = -p\).
• The product of the roots is \(\alpha \beta = q\).

These relationships are insightful because they give us a way to work with roots without explicitly solving the entire quadratic equation for \(\alpha\) and \(\beta\). For example, when you know \(p\) and \(q\), you can find the sum and product of the roots directly. This makes Viète's Formulas incredibly handy in more complex algebraic manipulations where solving directly might be cumbersome. In this exercise, we used Viète's Formulas to substitute values and make calculations simpler for finding an equation with \(\frac{\alpha}{\beta}\) as a root.

Roots of Equations

The roots of an equation are the values that make the equation true when substituted into it. For a quadratic equation \(x^2 + px + q = 0\), the roots \(\alpha\) and \(\beta\) are the solutions for \(x\). To solve for these, you typically apply the quadratic formula, but sometimes, you can work with properties of the roots themselves using Viète's Formulas.

However, when the problem is about a function of the roots, like \(\frac{\alpha}{\beta}\), you need to go beyond just finding \(\alpha\) and \(\beta\). Here, our task was to determine another equation in which this fraction would be a root. This involves using the known roots to derive new relationships and manipulate expressions algebraically to form the desired equation.

Understanding the nature of roots and how they relate to the coefficients through Viète's Formulas allows you to explore deeper algebraic manipulation to come up with such novel equations.

Algebraic Manipulation

Algebraic manipulation is the process of rearranging and simplifying expressions using mathematical operations. It involves altering an equation to make it easier to solve or understand without changing its solutions. In this exercise, we used algebraic manipulation to transform our expressions and find a fitting equation where \(\frac{\alpha}{\beta}\) is a root.

• We started by substituting \(y = \frac{\alpha}{\beta}\) and reconsidered one of \(\alpha\) as \(y\beta\).
• Then, using the relationships from Viète's Formulas, specifically \(\alpha\beta = q\) and \(\alpha + \beta = -p\), we substituted back into the expression.
• Finally, this substitution led to the quadratic equation \(qy^2 + (2q - p^2)y + q = 0\).

This manipulation was crucial to rearranging the terms effectively, allowing us to match one of the given choices. Mastering algebraic manipulation can significantly reduce the complexity and effort required to find desired results in equations involving roots or complex functions of roots.