If `alpha, beta` are roots of the equation `ax^2 + bx + c = 0` then the equation whose roots are `2alpha + 3beta` and `3alpha + 2beta` is

Options

A. `ab x^(2) - (a+b) cx + (a+b)^(2) = 0`
B. `ac x^(2) -(a+c) bx + (a + c)^(2) = 0`
C. `ac x^(2) + (a+c) bx -(a+c) bx -(a+c)^(2) = 0`
D. none of these

Answer

To find the equation whose roots are \(2\alpha + 3\beta\) and \(3\alpha + 2\beta\), we start with the given quadratic equation:

\[ ax^2 + bx + c = 0 \]

where \(\alpha\) and \(\beta\) are the roots. From Vieta's formulas, we know:

1. The sum of the roots \(\alpha + \beta = -\frac{b}{a}\)
2. The product of the roots \(\alpha \beta = \frac{c}{a}\)

### Step 1: Calculate the new roots

Let:
\[ A = 2\alpha + 3\beta \]
\[ B = 3\alpha + 2\beta \]

### Step 2: Find the sum of the new roots

The sum of the new roots \(A + B\) is:
\[ A + B = (2\alpha + 3\beta) + (3\alpha + 2\beta) = 5\alpha + 5\beta = 5(\alpha + \beta) \]

Substituting the value of \(\alpha + \beta\):
\[ A + B = 5\left(-\frac{b}{a}\right) = -\frac{5b}{a} \]

### Step 3: Find the product of the new roots

The product of the new roots \(A \cdot B\) is:
\[ A \cdot B = (2\alpha + 3\beta)(3\alpha + 2\beta) \]

Expanding this:
\[ A \cdot B = 6\alpha^2 + 4\alpha\beta + 9\alpha\beta + 6\beta^2 = 6\alpha^2 + 13\alpha\beta + 6\beta^2 \]

Now, we can express \(\alpha^2 + \beta^2\) in terms of \(\alpha + \beta\) and \(\alpha \beta\):
\[ \alpha^2 + \beta^2 = (\alpha + \beta)^2 - 2\alpha\beta = \left(-\frac{b}{a}\right)^2 - 2\left(\frac{c}{a}\right) = \frac{b^2}{a^2} - \frac{2c}{a} \]

Thus:
\[ A \cdot B = 6\left(\frac{b^2}{a^2} - \frac{2c}{a}\right) + 13\left(\frac{c}{a}\right) \]
\[ = \frac{6b^2}{a^2} - \frac{12c}{a} + \frac{13c}{a} = \frac{6b^2}{a^2} + \frac{c}{a} \]

### Step 4: Form the new quadratic equation

The new quadratic equation with roots \(A\) and \(B\) can be expressed as:
\[ x^2 - (A + B)x + (A \cdot B) = 0 \]

Substituting the values we found:
\[ x^2 - \left(-\frac{5b}{a}\right)x + \left(\frac{6b^2}{a^2} + \frac{c}{a}\right) = 0 \]

Multiplying through by \(a^2\) to eliminate the denominators:
\[ a^2x^2 + 5abx + (6b^2 + ac) = 0 \]

### Final Result

Thus, the required equation whose roots are \(2\alpha + 3\beta\) and \(3\alpha + 2\beta\) is:
\[ a^2x^2 + 5abx + (6b^2 + ac) = 0 \]

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