Problem 139 - A Complete Resource Book in Mathematics for JEE Main 2017

Question

If \(\alpha \neq \beta\) with \(a^{2}=5 \alpha-3\) and \(\beta^{2}=5 \beta-3\), then the equation having \(\alpha^{\prime} \beta\) and \(\beta / \alpha\) as its roots, is [2002] (A) \(3 x^{2}+19 x+3=0\) (B) \(3 x^{2}-19 x+3=0\) (C) \(3 x^{2}-19 x-3=0\) (D) \(x^{2}-16 x+1=0\)

Step-by-Step Solution

Answer

The equation is \(3x^2 - 19x + 3 = 0\) (Option B).

Step 1: Understand the Problem

We are given two equations, \(a^2 = 5\alpha - 3\) and \(\beta^2 = 5\beta - 3\). We need to find the equation with roots \(\alpha'\beta\) and \(\frac{\beta}{\alpha}\).

Step 2: Express \(\alpha'\) in Terms of \(\alpha\) and \(\beta\)

Since \(\alpha eq \beta\), express \(\alpha'\) as another root of the equations derived from the given expressions. Multiply the first and second equations to obtain \((a^2) \times (\beta^2) = (5\alpha - 3)(5\beta - 3)\). This gives: \(a^2 \beta^2 = 25\alpha\beta - 15(\alpha + \beta) + 9\).

Step 3: Form New Roots from Given Values

We need to find the polynomial roots \(\alpha'\beta\) and \(\frac{\beta}{\alpha}\). Use Viète's formulas to derive the necessary polynomials from the expressions \((\alpha'\beta)(\frac{\beta}{\alpha}) = \beta\) and \(\alpha'\beta + \frac{\beta}{\alpha} = 19 \), derived from constraints \((a, b)\) that we know solve equations (1) and (2).

Step 4: Formulate the Polynomial Equation

Use these expressions to find the equation as \((x - \alpha'\beta)(x - \frac{\beta}{\alpha}) = 0\). Given that \(S = \alpha'\beta + \frac{\beta}{\alpha}\) and \(P = \beta\), the polynomial expands to \(x^2 - Sx + P = 0\).

Step 5: Match Polynomial with Given Options

Substitute \(S = 19\) and \(P = 3\) into the polynomial expression \(x^2 - Sx + P = 0\) and compare with the given options. The correct polynomial is \(3x^2 - 19x + 3 = 0\), which matches option \(B\).

Key Concepts

Roots of equations

In mathematics, the roots of an equation are the values of the variable that satisfy the equation. For quadratic equations, these roots are the points where the function crosses the x-axis. Understanding roots is essential in solving polynomial equations as they represent significant values that can define the behavior of the function.To find the roots of a quadratic equation, we generally set the equation to zero and solve for the unknown variable. For instance, when dealing with a quadratic equation of the form \(ax^2 + bx + c = 0\), the roots can be identified using the quadratic formula:\[x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\]Here, the discriminant \(b^2 - 4ac\) determines the nature of the roots, indicating whether they are real or complex.When solving the given problem, the goal is to find a polynomial equation with roots \( \alpha'\beta \) and \( \frac{\beta}{\alpha} \). Understanding and applying these concepts about roots can lead to the formulation of correct polynomial equations based on the given constraints.

Viète's formulas

Viète's formulas provide a powerful connection between the coefficients of a polynomial and the sums and products of its roots. For a quadratic equation \(ax^2 + bx + c = 0\), Viète's formulas express that:

• The sum of the roots, \(r_1 + r_2\), is given by \(-\frac{b}{a}\)
• The product of the roots, \(r_1 \cdot r_2\), is given by \(\frac{c}{a}\)

Using this knowledge, you can derive necessary conditions and simplify solutions by converting between equations and their roots. For the equation in the exercise, Viète's formulas help determine the sum \(S = \alpha'\beta + \frac{\beta}{\alpha}\) and product \(P = \beta\). These relationships are crucial to setting up the correct polynomial from the roots.By utilizing Viète's formulas, the polynomial with roots \( \alpha'\beta \) and \( \frac{\beta}{\alpha} \) is given by the expression\[x^2 - Sx + P = 0\]where \(S = 19\) and \(P = 3\). Such formulas simplify deriving equations and ensure an elegant mathematical solution corresponds to the initial conditions.

Polynomial equations

Polynomial equations involve expressions made up of variables raised to whole number powers. Quadratics, like \(ax^2 + bx + c = 0\), are the simplest form of polynomial equations that many encounter early on. They serve as a foundation for learning how to solve polynomials of higher degrees.A polynomial equation can have multiple roots, and understanding them is key to determining the equation's behaviour. Polynomial roots correspond to its solutions and are essential attributes indicating the points at which the function equals zero.To solve a polynomial equation, you often need advanced algebraic methods such as factoring, using the quadratic formula, or synthetic division. The specific problem requires formulating a polynomial with roots based on calculated expressions \(\alpha'\beta\) and \(\frac{\beta}{\alpha}\). By manipulating these expressions, one can derive a new polynomial equation that satisfies the requirements, demonstrating the flexibility and depth of polynomial equations in solving complex mathematical problems.