the quardatic equation whose roots are squaresof the sum and of the difference of the roots of 2x2 + 2(m+n)x2 + m2 + n2 = 0

To solve the problem of finding the quadratic equation whose roots are the squares of the sum and the difference of the roots of the equation \(2x^2 + 2(m+n)x + m^2 + n^2 = 0\), let's break it down step by step.

### Step 1: Find the roots of the given quadratic equation
The given quadratic equation is:
\[
2x^2 + 2(m+n)x + (m^2 + n^2) = 0
\]

First, we divide the entire equation by 2 to simplify it:
\[
x^2 + (m+n)x + \frac{m^2 + n^2}{2} = 0
\]

Let's denote the roots of this equation as \( \alpha \) and \( \beta \).

### Step 2: Use Vieta's formulas
According to Vieta's formulas for a quadratic equation \(x^2 + px + q = 0\):
- The sum of the roots \( \alpha + \beta = -p \)
- The product of the roots \( \alpha \beta = q \)

In our case:
\[
\alpha + \beta = -(m+n)
\]
\[
\alpha \beta = \frac{m^2 + n^2}{2}
\]

### Step 3: Find the roots of the new quadratic equation
We are asked to find the quadratic equation whose roots are the squares of the sum and the difference of the roots of the original equation. Therefore, the roots of the new equation will be:

1. \( (\alpha + \beta)^2 = (-(m+n))^2 = (m+n)^2 \)
2. \( (\alpha - \beta)^2 \)

To find \( (\alpha - \beta)^2 \), we use the identity:
\[
(\alpha - \beta)^2 = (\alpha + \beta)^2 - 4\alpha\beta
\]

Substituting the known values:
\[
(\alpha - \beta)^2 = (m+n)^2 - 4 \times \frac{m^2 + n^2}{2} 
\]
\[
(\alpha - \beta)^2 = (m+n)^2 - 2(m^2 + n^2)
\]
\[
(\alpha - \beta)^2 = m^2 + 2mn + n^2 - 2m^2 - 2n^2 
\]
\[
(\alpha - \beta)^2 = -m^2 + 2mn - n^2
\]
\[
(\alpha - \beta)^2 = -(m-n)^2
\]

### Step 4: Form the new quadratic equation
The roots of the new quadratic equation are \( (m+n)^2 \) and \( -(m-n)^2 \). Therefore, the quadratic equation with these roots is:

\[
x^2 - \left[(m+n)^2 + -(m-n)^2 \right]x + (m+n)^2 \times -(m-n)^2 = 0
\]

Simplifying the terms:
1. The sum of the roots:
\[
(m+n)^2 + (m-n)^2
\]
Expanding each:
\[
= m^2 + 2mn + n^2 + m^2 - 2mn + n^2
\]
\[
= 2m^2 + 2n^2
\]
2. The product of the roots:
\[
(m+n)^2 \times -(m-n)^2 = -(m^2 - n^2)^2
\]

Thus, the quadratic equation is:
\[
x^2 - 2(m^2 + n^2)x - (m^2 - n^2)^2 = 0
\]

### Final Answer:
The quadratic equation whose roots are the squares of the sum and the difference of the roots of the original equation is:
\[
\boxed{x^2 - 2(m^2 + n^2)x - (m^2 - n^2)^2 = 0}
\]

Would you like more details, or do you have any other questions?

### Relative Questions
1. What are the roots of a quadratic equation and how are they found?
2. How does Vieta's formulas relate to the sum and product of roots?
3. What is the significance of the discriminant in a quadratic equation?
4. How can we derive the quadratic formula from a general quadratic equation?
5. How does one identify the nature of the roots of a quadratic equation?
6. Can the roots of a quadratic equation be complex? If so, how does it affect the quadratic formula?
7. What happens if the coefficient of the \(x^2\) term in a quadratic equation is zero?
8. How can you derive a quadratic equation if only the roots are known?

### Tip:
Always check the discriminant \(b^2 - 4ac\) of a quadratic equation to determine the nature of the roots (real and distinct, real and equal, or complex).

Explore how to find the quadratic equation whose roots are the squares of the sum and difference of the roots of 2x^2 + 2(m+n)x + m^2 + n^2 = 0. Detailed step-by-step solution and derivation included.

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