Problem 22 - Algebra 1 Course in Mathematics for the IIT-JEE and Other Engineering Exams

Question

If \(\alpha\) and \(\beta\) are roots of the equation \(A x^{2}+B x\) \(+C=0\), then value of \(a^{3}+\beta^{3}\) is [RPET-1996; DCE - 2005] (a) \(\frac{3 A B C-B^{3}}{A^{3}}\) (b) \(\frac{3 A B C+B^{3}}{A^{3}}\) (c) \(\frac{B^{3}-3 A B C}{A^{3}}\) (d) \(\frac{B^{3}-3 A B C}{B^{3}}\)

Step-by-Step Solution

Answer

The value of \(\alpha^3 + \beta^3\) is \(\frac{B^3 - 3ABC}{A^3}\) (Option c).

Step 1: Recall the Quadratic Formula and Properties

For a quadratic equation of the form \(Ax^2 + Bx + C = 0\), the sum of the roots (\(\alpha + \beta\)) is given by \(-\frac{B}{A}\), and the product of the roots (\(\alpha \beta\)) is given by \(\frac{C}{A}\). These relationships come from Viète's formulas.

Step 2: Use Identities to Relate \(\alpha^3 + \beta^3\)

Based on the identities from algebra, for two numbers \(\alpha\) and \(\beta\), \(\alpha^3 + \beta^3\) can be expressed as \((\alpha + \beta)(\alpha^2 - \alpha \beta + \beta^2)\). This can be further broken down using \(\alpha \beta\) and \(\alpha + \beta\).

Step 3: Simplify Using Known Sum and Product of Roots

First, calculate \(\alpha^2 - \alpha \beta + \beta^2\) using the identity \[(\alpha + \beta)^2 = \alpha^2 + 2\alpha \beta + \beta^2\].\Rearranging gives \(\alpha^2 + \beta^2 = (\alpha + \beta)^2 - 2\alpha \beta\).\Thus, \(\alpha^2 - \alpha \beta + \beta^2 = (\alpha^2 + \beta^2) - \alpha \beta\).

Step 4: Substitute Known Values

Substitute \(\alpha + \beta = -\frac{B}{A}\) and \(\alpha \beta = \frac{C}{A}\) into the expression from Step 3:\(\alpha^3 + \beta^3 = (\alpha + \beta)((\alpha + \beta)^2 - 3\alpha \beta)\).This becomes:\[\alpha^3 + \beta^3 = \left(-\frac{B}{A}\right) \left(\left(-\frac{B}{A}\right)^2 - 3\left(\frac{C}{A}\right)\right)\].

Step 5: Simplify the Expression

Simplify the equation:\[\left(-\frac{B}{A}\right)^2 = \frac{B^2}{A^2}\]\(\alpha^3 + \beta^3 = \left(-\frac{B}{A}\right) \left(\frac{B^2}{A^2} - \frac{3C}{A}\right)\).Simplify further to get\[= \left(-\frac{B}{A}\right) \left(\frac{B^2 - 3AC}{A^2}\right) = \frac{B^3 - 3ABC}{A^3}\].

Step 6: Compare with Given Options

The result obtained is \(\frac{B^3 - 3ABC}{A^3}\). Compare this with the given options. The correct option is (c) \(\frac{B^3 - 3ABC}{A^3}\).

Key Concepts

Roots of Equations

Understanding the roots of quadratic equations is crucial in solving these equations. A quadratic equation is typically expressed as \(Ax^2 + Bx + C = 0\). The roots of such an equation, often represented by \(\alpha\) and \(\beta\), are the values of \(x\) that satisfy the equation. In other words, if you plug these values back into the equation, the left-hand side equals zero.

The roots can be found using the quadratic formula:

• \(x = \frac{-B \pm \sqrt{B^2 - 4AC}}{2A}\)

This formula stems from completing the square method, and it helps in computing the exact values of \(\alpha\) and \(\beta\) without needing to factor the equation.

Understanding the relationship between the coefficients and the roots gives insight into the nature and properties of the quadratic equation. For instance, the discriminant \(B^2 - 4AC\) indicates whether the roots are real and distinct, real and identical, or complex conjugates.

Viete's Formulas

Viète's formulas provide a bridge between the roots \(\alpha\) and \(\beta\) of a polynomial equation and its coefficients. For a quadratic equation \(Ax^2 + Bx + C = 0\), these formulas are expressed as:

• Sum of the roots: \(\alpha + \beta = -\frac{B}{A}\)
• Product of the roots: \(\alpha \beta = \frac{C}{A}\)

These equations illustrate a direct relationship between the arithmetic operations on the roots and the coefficients of the polynomial. This neat relationship is rooted in the way polynomials are structured and expanded.

Viète's formulas are not only essential in quickly evaluating expressions involving the roots but also in simplifying algebraic identities, such as the one required for solving \(\alpha^3 + \beta^3\). For example, the identity utilized during the solution, \((\alpha + \beta)(\alpha^2 - \alpha \beta + \beta^2)\), is handled more easily when Viète’s formulas are applied.

Algebraic Identities

Algebraic identities are equations that hold true for all values of the involved variables. They form foundational tools for manipulation and simplification of expressions. For example, in finding \(\alpha^3 + \beta^3\), the identity \((\alpha + \beta)(\alpha^2 - \alpha \beta + \beta^2)\) is employed.

To break down \(\alpha^3 + \beta^3\) using identities, notice the expression:

• \((\alpha + \beta)^2 = \alpha^2 + 2\alpha \beta + \beta^2\)
• From here we derive: \(\alpha^2 + \beta^2 = (\alpha + \beta)^2 - 2\alpha \beta\)

Using Viète's formulas, the identities are substituted with known values to simplify complex expressions. After substitution, the result shows how these algebraic identities elegantly simplify into the final form, demonstrating their power and utility in problem-solving.

These identities are integral to algebra as they enable the conversion of complex expressions into manageable forms, making algebraic manipulation and problem-solving more efficient.