Problem 30 - A Complete Resource Book in Mathematics for JEE Main 2017

Question

If \(\alpha, \beta\) be the roots of \(x^{2}+p x-q=0\) and \(\gamma, \delta\) be the roots of \(x^{2}+p x+r=0, q+r \neq 0\), then \(\frac{(\alpha-\gamma)(\alpha-\delta)}{(\beta-\gamma)(\beta-\delta)}=\) (A) 1 (B) \(q\) (C) \(\vec{r}\) (D) \(q+r\)

Step-by-Step Solution

Answer

The answer is 1, which corresponds to option (A).

Step 1: Use Vieta's Formulas

For the equation \(x^2 + px - q = 0\), Vieta's formulas give us \(\alpha + \beta = -p\) and \(\alpha \beta = -q\). Similarly, for \(x^2 + px + r = 0\), we have \(\gamma + \delta = -p\) and \(\gamma \delta = r\).

Step 2: Express Terms Using Root Relations

Rewrite \((\alpha - \gamma)(\alpha - \delta)\) and \((\beta - \gamma)(\beta - \delta)\) using the difference of roots: \[ (x - \gamma)(x - \delta) = x^2 - (\gamma + \delta)x + \gamma\delta \].Thus, substitute for \(x = \alpha\) and \(x = \beta\) respectively:\[ (\alpha - \gamma)(\alpha - \delta) = \alpha^2 - (\gamma + \delta)\alpha + \gamma\delta \],\[ (\beta - \gamma)(\beta - \delta) = \beta^2 - (\gamma + \delta)\beta + \gamma\delta \].

Step 3: Substitute Values from Vieta's Formulas

Since \(\gamma + \delta = -p\) and \(\gamma \delta = r\), we can substitute into these expressions:\[(\alpha - \gamma)(\alpha - \delta) = \alpha^2 + p\alpha + r \],\[(\beta - \gamma)(\beta - \delta) = \beta^2 + p\beta + r \].

Step 4: Evaluate \(\alpha^2 + p\alpha + r\) and \(\beta^2 + p\beta + r\)

Since \(\alpha\) and \(\beta\) are roots of \(x^2 + px - q = 0\), it follows that:\[ \alpha^2 + p\alpha = q \] and \[ \beta^2 + p\beta = q \].Therefore, substitute \(q\) for these expressions:\[(\alpha - \gamma)(\alpha - \delta) = q + r\],\[(\beta - \gamma)(\beta - \delta) = q + r\].

Step 5: Simplify the Expression

We now have that \[ \frac{(\alpha - \gamma)(\alpha - \delta)}{(\beta - \gamma)(\beta - \delta)} = \frac{q + r}{q + r} = 1 \].

Step 6: Conclusion: Final Result

The resultant value is 1, which corresponds to option (A).

Key Concepts

Vieta's Formulas

Vieta's Formulas are a helpful tool to deal with quadratic equations, especially when working with roots. For a quadratic equation in the form \( ax^2 + bx + c = 0 \), Vieta's Formulas allows us to find relationships between the coefficients of the equation and its roots, usually denoted as \( \alpha \) and \( \beta \). Here’s what Vieta's Formulas tell us:

• The sum of the roots \( \alpha + \beta \) is given by \( -\frac{b}{a} \).
• The product of the roots \( \alpha \beta \) is given by \( \frac{c}{a} \).

Using these formulas, it's much simpler to express and manipulate the roots without solving the equation directly. In our problem, for the equation \( x^2 + px - q = 0 \), we have \( \alpha + \beta = -p \) and \( \alpha \beta = -q \). Similarly, for \( x^2 + px + r = 0 \), we find \( \gamma + \delta = -p \) and \( \gamma \delta = r \).
This simplification can guide further algebraic manipulation when dealing with expressions involving these roots.

Roots of Quadratic Equation

The roots of a quadratic equation, such as \( x^2 + px = 0 \), are values of \( x \) that make the equation equal zero. In other words, they are where the graph of the equation intersects the x-axis. For any quadratic equation in the form \( ax^2 + bx + c = 0 \), the roots can be found using the quadratic formula:

• \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \)

These roots can be either real or complex numbers, and the expression inside the square root, \( b^2 - 4ac \) (known as the discriminant), helps determine the nature of the roots:

• If the discriminant is positive, there are two distinct real roots.
• If the discriminant is zero, there is exactly one real root (a double root).
• If the discriminant is negative, the roots are complex and come in conjugate pairs.

In the given exercise, understanding the roots \( \alpha, \beta \) for one equation and \( \gamma, \delta \) for another, helps us identify how expressions using these roots behave and simplify.

Algebraic Manipulation

Algebraic manipulation involves rewriting mathematical expressions in different forms to reveal insightful properties or to simplify calculations. For example, simplifying expressions involving roots of quadratic equations often requires skilled manipulation.
In our exercise, to solve \[ \frac{(\alpha - \gamma)(\alpha - \delta)}{(\beta - \gamma)(\beta - \delta)} \], we started by expressing terms like \( (\alpha - \gamma)(\alpha - \delta) \) using the relation \( (x - \gamma)(x - \delta) = x^2 - (\gamma + \delta)x + \gamma\delta \). By setting \( x \) as each of \( \alpha \) and \( \beta \), and substituting values from Vieta's Formulas, we identified connections between original expressions and derived ones.
This kind of manipulation is invaluable, particularly in transforming complex problems into manageable solutions. By aligning the original expressions with known formulas, we simplified \( \alpha^2 + p\alpha = q \) and \( \beta^2 + p\beta = q \), eventually allowing us to show that the expression simplifies to 1 by division.