In the quadratic equation `ax^2 + bx + c = 0`. if `delta = b^2-4ac` and `alpha+beta , alpha^2+beta^2 , alpha^3+beta^3` and `alpha,beta` are the roots of `ax^2 + bx + c =0`

Options

A. `Delta!=0`
B. `bDelta=0`
C. `cb!=0`
D. `Delta=0`

Explanation

To solve the problem, we need to analyze the given quadratic equation and the conditions provided. Let's break it down step by step.

### Step 1: Understand the Quadratic Equation
The standard form of a quadratic equation is given by:
\[ ax^2 + bx + c = 0 \]
where \( a \), \( b \), and \( c \) are constants.

### Step 2: Identify the Roots
Let \( \alpha \) and \( \beta \) be the roots of the quadratic equation. According to Vieta's formulas:
- The sum of the roots \( \alpha + \beta \) is given by:
\[ \alpha + \beta = -\frac{b}{a} \]
- The product of the roots \( \alpha \beta \) is given by:
\[ \alpha \beta = \frac{c}{a} \]

### Step 3: Calculate \( \alpha^2 + \beta^2 \) and \( \alpha^3 + \beta^3 \)
Using the identities for the sums of squares and cubes:
1. The sum of squares can be calculated as:
\[ \alpha^2 + \beta^2 = (\alpha + \beta)^2 - 2\alpha\beta = \left(-\frac{b}{a}\right)^2 - 2\left(\frac{c}{a}\right) = \frac{b^2}{a^2} - \frac{2c}{a} \]
2. The sum of cubes can be calculated as:
\[ \alpha^3 + \beta^3 = (\alpha + \beta)(\alpha^2 + \beta^2 - \alpha\beta) = \left(-\frac{b}{a}\right)\left(\frac{b^2}{a^2} - \frac{2c}{a} - \frac{c}{a}\right) \]
\[ = -\frac{b}{a}\left(\frac{b^2 - 3c}{a}\right) = -\frac{b(b^2 - 3c)}{a^2} \]

### Step 4: Set Up the Condition for Geometric Progression
It is given that \( \alpha + \beta \), \( \alpha^2 + \beta^2 \), and \( \alpha^3 + \beta^3 \) are in geometric progression (G.P.). For three numbers \( x, y, z \) to be in G.P., the following condition must hold:
\[ y^2 = xz \]
Substituting our expressions:
\[ \left(\alpha^2 + \beta^2\right)^2 = (\alpha + \beta)(\alpha^3 + \beta^3) \]

### Step 5: Substitute and Simplify
Substituting the expressions we derived:
1. Left-hand side:
\[ \left(\frac{b^2}{a^2} - \frac{2c}{a}\right)^2 = \left(\frac{b^2 - 2ac}{a^2}\right)^2 \]
2. Right-hand side:
\[ \left(-\frac{b}{a}\right)\left(-\frac{b(b^2 - 3c)}{a^2}\right) = \frac{b^2(b^2 - 3c)}{a^3} \]

Setting these equal gives:
\[ \left(\frac{b^2 - 2ac}{a^2}\right)^2 = \frac{b^2(b^2 - 3c)}{a^3} \]

### Step 6: Cross-Multiply and Rearrange
Cross-multiplying and simplifying leads to:
\[ (b^2 - 2ac)^2 = b^2(b^2 - 3c)a \]

### Step 7: Analyze the Result
After simplification, we arrive at:
\[ b^4 - 4ab^2c + 4a^2c^2 = b^4 - 3b^2ac \]
This simplifies to:
\[ -4ac + 3ac = 0 \]
Thus, we find:
\[ b^2 - 4ac = 0 \]

### Conclusion
This implies that the discriminant \( \Delta = b^2 - 4ac = 0 \). Therefore, the roots \( \alpha \) and \( \beta \) are equal, indicating that the quadratic has a double root.