If alpha - 2,beta - 2 are the roots of x^2 + 7x + 3 = 0 then 2alpha + 1,2beta + 1 will be the roots of x^2 + ax + b = 0 . The value of a + b=

Given that \(\alpha - 2\) and \(\beta - 2\) are the roots of the equation \(x^2 + 7x + 3 = 0\), we need to determine the values of \(a\) and \(b\) in the equation \(x^2 + ax + b = 0\) where \(2\alpha + 1\) and \(2\beta + 1\) are the roots, and then find \(a + b\).

### Step 1: Find \(\alpha\) and \(\beta\)
The roots \(\alpha - 2\) and \(\beta - 2\) satisfy the equation \(x^2 + 7x + 3 = 0\).

By Vieta's formulas:
- The sum of the roots is \((\alpha - 2) + (\beta - 2) = -7\)
- The product of the roots is \((\alpha - 2)(\beta - 2) = 3\)

This simplifies to:
\[
\alpha + \beta - 4 = -7 \quad \Rightarrow \quad \alpha + \beta = -3
\]
\[
(\alpha - 2)(\beta - 2) = 3 \quad \Rightarrow \quad \alpha\beta - 2(\alpha + \beta) + 4 = 3
\]
Substituting \(\alpha + \beta = -3\) into the equation:
\[
\alpha\beta + 6 + 4 = 3 \quad \Rightarrow \quad \alpha\beta = -7
\]

### Step 2: Determine the new roots \(2\alpha + 1\) and \(2\beta + 1\)
We know that the roots of the new equation are \(2\alpha + 1\) and \(2\beta + 1\).

Using Vieta's formulas for the new equation \(x^2 + ax + b = 0\):
- The sum of the roots is:
\[
(2\alpha + 1) + (2\beta + 1) = 2(\alpha + \beta) + 2 = 2(-3) + 2 = -6 + 2 = -4
\]
Thus, \(a = 4\).

- The product of the roots is:
\[
(2\alpha + 1)(2\beta + 1) = 4\alpha\beta + 2(\alpha + \beta) + 1 = 4(-7) + 2(-3) + 1 = -28 - 6 + 1 = -33
\]
Thus, \(b = -33\).

### Step 3: Calculate \(a + b\)
Finally, we find:
\[
a + b = 4 + (-33) = -29
\]

Thus, the value of \(a + b\) is \(\boxed{-29}\).

Would you like more details or have any questions?

Here are some related questions:
1. How do Vieta's formulas apply to quadratic equations?
2. What is the importance of knowing the sum and product of roots in a quadratic equation?
3. How can transformations of roots be used to derive new quadratic equations?
4. What are the implications of a negative discriminant in a quadratic equation?
5. How can the relationships between roots be used to solve more complex polynomial equations?

**Tip:** When dealing with transformed roots in quadratic equations, always start by applying Vieta's formulas to the original and transformed equations to simplify your calculations.

Explore how to calculate new roots of a quadratic equation based on transformed roots. Solve for coefficients using Vieta's formulas and understand the implications of root transformations. Suitable for advanced high school students and beyond.

Math Problem Statement

Solution

Ask a new question for Free

By Image

Math Problem Analysis

Mathematical Concepts

Formulas

Theorems

Suitable Grade Level

Related Recommendation