If `alpha, beta` are the roots of the equation `5x^(2)-5x+1=0`, then find the value of `(1)/(2)(a+b alpha+c alpha^(2)+d alpha^(3))+(1)/(2)(a+b beta+c beta^(2)+d beta^(3))`.

Answer

To solve the problem, we need to find the value of the expression:

\[ \frac{1}{2}\left(a + b\alpha + c\alpha^2 + d\alpha^3\right) + \frac{1}{2}\left(a + b\beta + c\beta^2 + d\beta^3\right) \]

where \(\alpha\) and \(\beta\) are the roots of the quadratic equation \(5x^2 - 5x + 1 = 0\).

### Step 1: Find the roots \(\alpha\) and \(\beta\)

Using the quadratic formula:

\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]

For our equation \(5x^2 - 5x + 1 = 0\), we have \(a = 5\), \(b = -5\), and \(c = 1\).

Calculating the discriminant:

\[ b^2 - 4ac = (-5)^2 - 4 \cdot 5 \cdot 1 = 25 - 20 = 5 \]

Now substituting into the quadratic formula:

\[ x = \frac{5 \pm \sqrt{5}}{10} = \frac{1 \pm \sqrt{5}/5}{2} \]

Thus, the roots are:

\[ \alpha = \frac{1 + \sqrt{5}/5}{2}, \quad \beta = \frac{1 - \sqrt{5}/5}{2} \]

### Step 2: Use Vieta's formulas

From Vieta's formulas, we know:

\[ \alpha + \beta = \frac{-(-5)}{5} = 1 \]
\[ \alpha \beta = \frac{1}{5} \]

### Step 3: Simplify the expression

We can factor out \(\frac{1}{2}\):

\[ \frac{1}{2}\left(2a + b(\alpha + \beta) + c(\alpha^2 + \beta^2) + d(\alpha^3 + \beta^3)\right) \]

Substituting \(\alpha + \beta = 1\):

\[ = \frac{1}{2}\left(2a + b \cdot 1 + c(\alpha^2 + \beta^2) + d(\alpha^3 + \beta^3)\right) \]

### Step 4: Calculate \(\alpha^2 + \beta^2\)

Using the identity:

\[ \alpha^2 + \beta^2 = (\alpha + \beta)^2 - 2\alpha\beta = 1^2 - 2 \cdot \frac{1}{5} = 1 - \frac{2}{5} = \frac{3}{5} \]

### Step 5: Calculate \(\alpha^3 + \beta^3\)

Using the identity:

\[ \alpha^3 + \beta^3 = (\alpha + \beta)(\alpha^2 + \beta^2 - \alpha\beta) = 1\left(\frac{3}{5} - \frac{1}{5}\right) = \frac{2}{5} \]

### Step 6: Substitute back into the expression

Now substituting back into our expression:

\[ = \frac{1}{2}\left(2a + b + c \cdot \frac{3}{5} + d \cdot \frac{2}{5}\right) \]

### Step 7: Final expression

Rearranging gives us:

\[ = \frac{1}{2}\left(2a + b + \frac{3c}{5} + \frac{2d}{5}\right) \]

### Step 8: Final result

Thus, the final value of the expression is:

\[ \frac{1}{2}\left(2a + b + \frac{3c}{5} + \frac{2d}{5}\right) \]