Problem 8 - Mathematics for IIT JEE Main and Advanced Differential Calculus Algebra Trigonometry

Question

If \(\alpha, \beta\) are roots of the equation \(6 x^{2}-6 x+1=0\), then find the value of \(\frac{1}{2}\left(a+b \alpha+c \alpha^{2}+d \alpha^{3}\right)+\frac{1}{2}\left(a+b \beta+c \beta^{2}+d \beta^{3}\right)\) in terms of \(a, b, c, d\).

Step-by-Step Solution

Answer

The value of the given expression in terms of a, b, c, and d is \(a + \frac{b}{2} + \frac{c}{3} + \frac{d}{6}(2α - 1)\).

Step 1: Identify the Sum and Product of Roots

Given the quadratic equation \(6x^2 - 6x + 1 = 0\), let's find the sum and the product of its roots, α and β, using the Viete's formulas: Sum of roots, \(α + β = -\frac{(-6)}{6} = 1\) Product of roots, \(αβ = \frac{1}{6}\)

Step 2: Compute the expression

We need to find the value of \(\frac{1}{2}(a + bα + cα^2 + dα^3) + \frac{1}{2}(a + bβ + cβ^2 + dβ^3)\) in terms of a, b, c, and d. Let's address each term separately and then sum the results.

Step 3: Calculate the terms with α

For the first term, \(\frac{1}{2}(a + bα + cα^2 + dα^3)\), calculate \(α^2\) and \(α^3\): \( α^2 = (α + β)^2 - 2αβ = 1^2 - 2\left(\frac{1}{6}\right) = \frac{1}{3} \) \( α^3 = α(α^2) = α\left(\frac{1}{3}\right) = \frac{α}{3} \) Now substitute these values into the first term: \( \frac{1}{2}(a + bα + cα^2 + dα^3) = \frac{1}{2}(a + bα + c\left(\frac{1}{3}\right) + d\left(\frac{α}{3}\right)) \)

Step 4: Calculate the terms with β

For the second term, \(\frac{1}{2}(a + bβ + cβ^2 + dβ^3)\), we use the relation \((1-\alpha) = \beta\) and do the same steps as above. Calculate \(\beta^2\) and \(\beta^3 \): \( β^2 = (α + β)^2 - 2αβ = 1^2 - 2\left(\frac{1}{6}\right) = \frac{1}{3} \) \( β^3 = β^2 \cdot β = (α + β - α^2) \cdot (1 - α) = \left(\frac{1}{3}\right)\left(\frac{-2}{6}\right) = -\frac{β}{3} \) Now substitute these values into the second term: \( \frac{1}{2}(a + bβ + cβ^2 + dβ^3) = \frac{1}{2}(a + b (1-\alpha) + c\left(\frac{1}{3}\right) + d\left(-\frac{β}{3}\right)) = \frac{1}{2}(a + b(1-\alpha) + c\left(\frac{1}{3}\right) - d\left(\frac{1-\alpha}{3}\right)) \)

Step 5: Combine and simplify

Combine the results from Steps 3 and 4 to find the value of the given expression: \( \frac{1}{2}(a + bα + cα^2 + dα^3) + \frac{1}{2}(a + bβ + cβ^2 + dβ^3) = \frac{1}{2}(a + bα + c\left(\frac{1}{3}\right) + d\left(\frac{α}{3}\right))+ \frac{1}{2}(a + b(1-\alpha) + c\left(\frac{1}{3}\right) - d\left(\frac{1-\alpha}{3}\right)) \) Simplify the expression: \( = \frac{1}{2}(2a + b(α+(1-\alpha)) + c\left(\frac{2}{3}\right) + d\left(\frac{α}{3} -\frac{1-\alpha}{3}\right)) \) = \(a + \frac{b}{2} + \frac{c}{3} + \frac{d}{6}(2α - 1)\)

Key Concepts

Roots of Equations

Understanding the roots of equations, especially quadratic ones, is key to solving and analyzing them. Roots are values that satisfy the equation such that when substituted into the equation, the result is zero. For example, for the equation \(6x^2 - 6x + 1 = 0\), the roots are \(\alpha\) and \(\beta\). These roots represent the x-values where the quadratic curve intersects the x-axis.
To find the roots of a quadratic equation, you can use formulas or factorization if applicable. An essential part of dealing with roots involves calculating the sum and product of these roots. Once you have the coefficients of an equation, the sum and product provide a way to express the relationship between the roots. This is neatly captured by Viete's formulas. These ideas help simplify expressions where roots are involved and allow for further insights through derived algebraic manipulations.

Viete's Formulas

Viete's formulas establish a fundamental link between the coefficients of a polynomial and its roots. When dealing with quadratic equations like \(ax^2 + bx + c = 0\), these formulas can be of great help. They tell us that the sum of the roots \(\alpha\) and \(\beta\) is equal to \(-\frac{b}{a}\), and the product of the roots is \(\frac{c}{a}\).
In the given equation \(6x^2 - 6x + 1 = 0\), Viete's Formulas allows us to quickly determine that:

• The sum of the roots \((\alpha + \beta) = 1\)
• The product of the roots \((\alpha \beta = \frac{1}{6})\)

By utilizing Viete's Formulas, calculations involving the roots become much easier. In our problem, we leveraged these to find higher powers of \(\alpha\) and \(\beta\), such as \(\alpha^2\) and \(\beta^2\), which simplifies handling algebraic expressions in exercises.

Algebraic Expressions

Handling algebraic expressions involving roots can appear daunting, but they often encapsulate many systematic approaches to make them simpler. In our given problem, we examined expressions of the form \(\frac{1}{2}(a + b\alpha + c\alpha^2 + d\alpha^3)\).
To deal with such expressions efficiently, one should substitute known values of \(\alpha\) and \(\beta\), and their powers. For example, \(\alpha^2\) can be derived using relationships like \((\alpha + \beta)^2 = \alpha^2 + 2\alpha\beta + \beta^2\). With known values of \(\alpha + \beta\) and \(\alpha\beta\), it becomes easier to simplify terms.
Breaking down expressions by substituting values from Viete's Formulas and simplifying each component separately allows you to solve complex algebraic expressions methodically. Always ensure consistency in terms and verify through substitution back into the original equation if possible.