If `alpha` and `beta` are roots of `ax^(2) + bx + c = 0`, then find the value of `a((alpha^(2) + beta^(2))/(beta alpha)) + b((alpha)/(beta) + (beta)/(alpha))`

Options

A. b
B. a
C. ab
D. a/b

Answer

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

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

where \(\alpha\) and \(\beta\) are the roots of the quadratic equation \(ax^2 + bx + c = 0\).

### Step 1: Express \(\alpha^2 + \beta^2\) in terms of \(\alpha + \beta\) and \(\alpha \beta\)

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

We know from Vieta's formulas:
\[ \alpha + \beta = -\frac{b}{a} \quad \text{and} \quad \alpha \beta = \frac{c}{a} \]

Substituting these into the identity:
\[ \alpha^2 + \beta^2 = \left(-\frac{b}{a}\right)^2 - 2\left(\frac{c}{a}\right) = \frac{b^2}{a^2} - \frac{2c}{a} \]

### Step 2: Substitute \(\alpha^2 + \beta^2\) into the expression

Now we substitute \(\alpha^2 + \beta^2\) into the original expression:
\[ a \left( \frac{\frac{b^2}{a^2} - \frac{2c}{a}}{\beta \alpha} \right) + b \left( \frac{\alpha}{\beta} + \frac{\beta}{\alpha} \right) \]

### Step 3: Simplify the first term

The first term becomes:
\[ a \left( \frac{\frac{b^2 - 2ac}{a^2}}{\frac{c}{a}} \right) = a \left( \frac{b^2 - 2ac}{ac} \right) = \frac{a(b^2 - 2ac)}{ac} = \frac{b^2 - 2ac}{c} \]

### Step 4: Simplify the second term

The second term can be simplified as follows:
\[ b \left( \frac{\alpha}{\beta} + \frac{\beta}{\alpha} \right) = b \left( \frac{\alpha^2 + \beta^2}{\alpha \beta} \right) \]
Using the earlier result for \(\alpha^2 + \beta^2\):
\[ = b \left( \frac{\frac{b^2}{a^2} - \frac{2c}{a}}{\frac{c}{a}} \right) = b \left( \frac{b^2 - 2ac}{c} \right) \]

### Step 5: Combine both terms

Now we combine both terms:
\[ \frac{b^2 - 2ac}{c} + \frac{b(b^2 - 2ac)}{c} = \frac{(b^2 - 2ac) + b(b^2 - 2ac)}{c} \]

Factoring out \((b^2 - 2ac)\):
\[ = \frac{(b^2 - 2ac)(1 + b)}{c} \]

### Final Answer

Thus, the value of the expression is:
\[ \frac{(b^2 - 2ac)(1 + b)}{c} \]