Problem 87 - A Complete Resource Book in Mathematics for JEE Main

Question

If the equations \(x^{2}-p x+q=0\) and \(x^{2}-a x+b=0\) have a common root and the other root of the second equation is the reciprocal of the other root of the first, then \((q-b)^{2}=\) (A) \(a q(p-b)^{2}\) (B) \(b q(p-a)^{2}\) (C) \(b q(p-b)^{2}\) (D) None of these

Step-by-Step Solution

Answer

(q-b)^2 = 0, which doesn't match any given option.

Step 1: Identify the Common Root

Let's assume the common root of both equations is \( \alpha \). For the first equation, using Vieta's formulas, we know \( \alpha + \beta = p \) and \( \alpha \beta = q \). For the second equation, we assume \( \alpha + \gamma = a \) and \( \alpha \gamma = b \).

Step 2: Relate the Roots Using Given Conditions

We are given that the other root of the second equation, \( \gamma \), is the reciprocal of the other root of the first equation, \( \beta \). Therefore, \( \gamma = \frac{1}{\beta} \).

Step 3: Express \(\beta\) and \(\gamma\) in Terms of Known Values

From the first equation, \( \beta \) can be expressed as \( \beta = p - \alpha \), and from the second equation, \( \gamma = a - \alpha \). Since \( \gamma = \frac{1}{\beta} \), equate \(a - \alpha = \frac{1}{p - \alpha}\).

Step 4: Solve the Root Relations

Given \( \alpha \gamma = b \), substitute \( \gamma = \frac{1}{\beta} \) into \( \alpha \frac{1}{\beta} = b \), we have \( \beta \alpha = \frac{1}{b} \). Thus, \( \beta = \frac{b}{\alpha}\). Substitute back to find \( q \).

Step 5: Substitute and Simplify the Expression for \((q-b)^{2}\)

From the relation \( \alpha \beta = q \), substitute \( \beta = \frac{b}{\alpha} \) to find \( q = b \). Now substitute \( q = b \) to simplify \((q-b)^{2}\). Thus, \((q-b)^{2} = 0\).

Step 6: Conclusion and Verification

Upon verification of all conditions and simplifications, it seems none of the provided options directly equal \((q-b)^{2}\). Thus, the answer for \((q-b)^{2}\) is 0, which is not presented as an option.

Key Concepts

Vieta's Formulas

Vieta's formulas are a helpful tool when dealing with quadratic equations. They allow you to connect the coefficients of the polynomial to the sum and product of its roots.
For a quadratic equation of the form \(x^2 - px + q = 0\), Vieta's formulas tell us that the sum of the roots (\(\alpha + \beta\)) is equal to \(p\), and the product of the roots (\(\alpha \beta\)) is equal to \(q\).

In practical terms, this means:

• Roots Sum: \(\alpha + \beta = p\)
• Roots Product: \(\alpha \beta = q\)

These expressions provide a quick way to find relationships between the roots without solving the full quadratic equation. This is especially useful when dealing with problems involving a common root or reciprocal roots across two quadratic equations.

Root Relationships

Understanding root relationships is crucial when you have two quadratic equations with a common root.
In the problem you are tackling, if one root is common, it allows us to set up specific equations to find other root relationships.

Let's assume \(\alpha\) is the common root:

• First Equation: \(\alpha + \beta = p\), \(\alpha \beta = q\)
• Second Equation: \(\alpha + \gamma = a\), \(\alpha \gamma = b\)

These relationships form the basis of solving the equations. By setting these conditions, we understand how one root influences the relationships in both equations. And, knowing one common element can guide you toward other unknown root values.
This is where reciprocal roots and other relationships become prominent, solving for the unknowns based on these initial root relationships.

Reciprocal Roots

Reciprocal roots introduce an interesting dynamic in understanding quadratic equations. If a root \(\gamma\) is said to be the reciprocal of another root \(\beta\), it means that \(\gamma = \frac{1}{\beta}\).
This relationship simplifies many calculations, especially when used in conjunction with Vieta's formulas.

Consider the equation:

• If \(\beta\) is a root of the first equation, then \(\beta = p - \alpha\)
• Thus its reciprocal \(\gamma = \frac{1}{\beta}\)
• Accordingly in the second equation where \(\gamma = a - \alpha\), this sets up an equation: \(a - \alpha = \frac{1}{\beta}\)

This makes calculations more straightforward by providing a direct relationship between \(\beta\) and \(\gamma\). Reciprocal roots offer a simplified path to solving root-related questions, giving a clear algebraic expression to guide the problem-solving strategy.

Common Root Conditions

When two quadratic equations share a common root, specific mathematical conditions must be satisfied for consistency. This commonly shared root forms a bridge that provides additional equations to solve for unknowns in each equation.

Given two equations:

• Equation 1: \(x^2 - px + q = 0\)
• Equation 2: \(x^2 - ax + b = 0\)

If \(\alpha\) is their common root:

• It appears in both equations, generating new expressions based on Vieta's relationships.
• You must ensure the non-common roots maintain a valid mathematical relationship, often through reciprocal or dependent relationships as outlined in the problem statement.

The importance of common roots lies in how they anchor polynomial relationships, guiding the equation's overall solution path. They become a pivotal part of understanding interconnected systems of equations.