Given that `alpha and beta` are the roots of the equation `x^(2)-x+7=0`, find
(i) The numerical value of `(alpha)/(beta+3)+(beta)/(alpha+3)`,
(ii) an equation whose roots are `(alpha)/(beta+3) and (beta)/(alpha+3)`.

Answer

To solve the problem step by step, we will break down the solution into two parts as given in the question.

### Given:
The quadratic equation is:
\[ x^2 - x + 7 = 0 \]

### Step 1: Find the roots (α and β)
Using Vieta's formulas:
- The sum of the roots \( \alpha + \beta = -\frac{b}{a} = -\frac{-1}{1} = 1 \)
- The product of the roots \( \alpha \beta = \frac{c}{a} = \frac{7}{1} = 7 \)

### Step 2: Calculate \( \frac{\alpha}{\beta + 3} + \frac{\beta}{\alpha + 3} \)
To find this expression, we will first find a common denominator:
\[ \frac{\alpha}{\beta + 3} + \frac{\beta}{\alpha + 3} = \frac{\alpha(\alpha + 3) + \beta(\beta + 3)}{(\beta + 3)(\alpha + 3)} \]

Now, simplify the numerator:
\[ \alpha(\alpha + 3) + \beta(\beta + 3) = \alpha^2 + 3\alpha + \beta^2 + 3\beta = (\alpha^2 + \beta^2) + 3(\alpha + \beta) \]

Using the identity \( \alpha^2 + \beta^2 = (\alpha + \beta)^2 - 2\alpha\beta \):
\[ \alpha^2 + \beta^2 = (1)^2 - 2(7) = 1 - 14 = -13 \]

Thus, we have:
\[ \alpha^2 + \beta^2 + 3(\alpha + \beta) = -13 + 3(1) = -13 + 3 = -10 \]

Now, substituting back into the expression:
\[ \frac{\alpha}{\beta + 3} + \frac{\beta}{\alpha + 3} = \frac{-10}{(\beta + 3)(\alpha + 3)} \]

### Step 3: Calculate \( (\beta + 3)(\alpha + 3) \)
Expanding this:
\[ (\beta + 3)(\alpha + 3) = \alpha\beta + 3\alpha + 3\beta + 9 = 7 + 3(1) + 9 = 7 + 3 + 9 = 19 \]

### Step 4: Final Calculation
Now substituting back:
\[ \frac{\alpha}{\beta + 3} + \frac{\beta}{\alpha + 3} = \frac{-10}{19} \]

### Answer for Part (i):
\[ \frac{\alpha}{\beta + 3} + \frac{\beta}{\alpha + 3} = -\frac{10}{19} \]

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### Part (ii): Find an equation whose roots are \( \frac{\alpha}{\beta + 3} \) and \( \frac{\beta}{\alpha + 3} \)

#### Step 1: Sum of the roots
From part (i), we know:
\[ \text{Sum} = \frac{\alpha}{\beta + 3} + \frac{\beta}{\alpha + 3} = -\frac{10}{19} \]

#### Step 2: Product of the roots
Calculating the product:
\[ \frac{\alpha}{\beta + 3} \cdot \frac{\beta}{\alpha + 3} = \frac{\alpha\beta}{(\beta + 3)(\alpha + 3)} = \frac{7}{19} \]

#### Step 3: Form the quadratic equation
Using the standard form \( x^2 - (\text{Sum})x + \text{Product} = 0 \):
\[ x^2 + \frac{10}{19}x + \frac{7}{19} = 0 \]

Multiplying through by 19 to eliminate the fraction:
\[ 19x^2 + 10x + 7 = 0 \]

### Answer for Part (ii):
The required quadratic equation is:
\[ 19x^2 + 10x + 7 = 0 \]

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