Key Concepts and Formulas

• Roots of a Quadratic Equation: For a quadratic equation $ax^2 + bx + c = 0$, the sum of the roots is $-\frac{b}{a}$ and the product of the roots is $\frac{c}{a}$.
• Arithmetic Progression (AP): Three terms $x, y, z$ are in AP if $2y = x + z$.
• Arithmetic-Geometric Progression (AGP): Three terms $x, y, z$ are considered to be in AGP if they satisfy a particular relationship. One common condition related to the given problem is that $2z = y + \frac{1}{x^2}$ for $x, y, z$ that are reciprocals.

Step-by-Step Solution

Step 1: Define the Roots and Apply Vieta's Formulas

• Why: Establish the relationship between the coefficients of the quadratic equation and its roots using Vieta's formulas. This is the foundation for connecting the given condition to $a, b, c$.
• Let $\alpha$ and $\beta$ be the roots of the quadratic equation $ax^2 + bx + c = 0$. Then, according to Vieta's formulas:
  • Sum of roots: $\alpha + \beta = -\frac{b}{a}$
  • Product of roots: $\alpha\beta = \frac{c}{a}$

Step 2: Express the Given Condition Algebraically

• Why: Translate the verbal statement "the sum of the roots is equal to the sum of the squares of their reciprocals" into a mathematical equation.
• The given condition is: $\alpha + \beta = \frac{1}{\alpha^2} + \frac{1}{\beta^2}$

Step 3: Simplify the Right-Hand Side

• Why: Simplify the right-hand side of the equation to express it in terms of the sum and product of the roots, which are known in terms of $a, b, c$.
• Combine the fractions and express the numerator in terms of $(\alpha + \beta)$ and $(\alpha\beta)$: $\frac{1}{\alpha^2} + \frac{1}{\beta^2} = \frac{\alpha^2 + \beta^2}{\alpha^2\beta^2} = \frac{(\alpha + \beta)^2 - 2\alpha\beta}{(\alpha\beta)^2}$

Step 4: Substitute Vieta's Formulas into the Equation

• Why: Replace $\alpha + \beta$ and $\alpha\beta$ with their expressions in terms of $a, b, c$ to obtain an equation involving only the coefficients.
• Substitute $\alpha + \beta = -\frac{b}{a}$ and $\alpha\beta = \frac{c}{a}$ into the equation: $-\frac{b}{a} = \frac{\left(-\frac{b}{a}\right)^2 - 2\left(\frac{c}{a}\right)}{\left(\frac{c}{a}\right)^2}$

Step 5: Simplify the Algebraic Expression

• Why: Simplify the equation to find a relationship between $a, b,$ and $c$.
• Simplify the expression: $-\frac{b}{a} = \frac{\frac{b^2}{a^2} - \frac{2c}{a}}{\frac{c^2}{a^2}} = \frac{\frac{b^2 - 2ac}{a^2}}{\frac{c^2}{a^2}} = \frac{b^2 - 2ac}{c^2}$ Cross-multiply: $-bc^2 = ab^2 - 2a^2c$ Rearrange the terms: $2a^2c = ab^2 + bc^2$

Step 6: Analyze the Relationship for AGP

• Why: Manipulate the equation to reveal the relationship between $a/c, b/a, c/b$ that indicates they are in AGP.
• Divide the equation $2a^2c = ab^2 + bc^2$ by $a^2b$: $\frac{2a^2c}{a^2b} = \frac{ab^2}{a^2b} + \frac{bc^2}{a^2b}$ $\frac{2c}{b} = \frac{b}{a} + \frac{c^2}{a^2}$
• Rewrite this equation as: $2\left(\frac{c}{b}\right) = \frac{b}{a} + \left(\frac{c}{a}\right)^2$

Step 7: Relate the Equation to the AGP Condition

• Why: Recognize that the derived equation matches the condition for the terms to be in AGP.
• Let $x = \frac{a}{c}, y = \frac{b}{a}, z = \frac{c}{b}$. Then $\frac{c}{a} = \frac{1}{x}$. The equation becomes $2z = y + \frac{1}{x^2}$ This form matches the required condition for the Arithmetic-Geometric Progression

Common Mistakes & Tips

• Careless Algebra: Double-check algebraic manipulations, especially when simplifying fractions and substituting values.
• Incorrect Progression Identification: Make sure you have a firm grasp of AP, GP, and HP definitions. The condition for AGP in this specific context is key.
• Forgetting Vieta's Formulas: These formulas are fundamental to solving problems involving roots and coefficients of polynomials.

Summary By using Vieta's formulas to relate the roots to the coefficients, and then simplifying the given condition, we arrive at the relationship $2\frac{c}{b} = \frac{b}{a} + \left(\frac{c}{a}\right)^2$. This can be rewritten as $2z = y + \frac{1}{x^2}$, where $x=a/c, y=b/a, z=c/b$, which is the condition for $a/c, b/a, c/b$ to be in Arithmetic-Geometric Progression.

Final Answer The final answer is \boxed{A}, which corresponds to option (A).