If `alpha, beta` are the roots of the quadratic equation `ax^(2)+bx+c=0,` form a quadratic equation whose roots are `alpha^(2)+beta^(2) and alpha^(-2) + beta^(-2)`.

Answer

To form a quadratic equation whose roots are \( \alpha^2 + \beta^2 \) and \( \alpha^{-2} + \beta^{-2} \), we will follow these steps:

### Step 1: Find \( \alpha^2 + \beta^2 \)

We know that:
\[ \alpha^2 + \beta^2 = (\alpha + \beta)^2 - 2\alpha\beta \]
From the given quadratic equation \( ax^2 + bx + c = 0 \), we have:
\[ \alpha + \beta = -\frac{b}{a} \quad \text{and} \quad \alpha\beta = \frac{c}{a} \]
Substituting these values:
\[ \alpha^2 + \beta^2 = \left(-\frac{b}{a}\right)^2 - 2\left(\frac{c}{a}\right) = \frac{b^2}{a^2} - \frac{2c}{a} = \frac{b^2 - 2ac}{a^2} \]

### Step 2: Find \( \alpha^{-2} + \beta^{-2} \)

Using the identity:
\[ \alpha^{-2} + \beta^{-2} = \frac{1}{\alpha^2} + \frac{1}{\beta^2} = \frac{\beta^2 + \alpha^2}{\alpha^2 \beta^2} \]
We already found \( \alpha^2 + \beta^2 \) in Step 1. Now, we need \( \alpha^2 \beta^2 \):
\[ \alpha^2 \beta^2 = (\alpha \beta)^2 = \left(\frac{c}{a}\right)^2 = \frac{c^2}{a^2} \]
Thus:
\[ \alpha^{-2} + \beta^{-2} = \frac{\alpha^2 + \beta^2}{\alpha^2 \beta^2} = \frac{\frac{b^2 - 2ac}{a^2}}{\frac{c^2}{a^2}} = \frac{b^2 - 2ac}{c^2} \]

### Step 3: Sum of the Roots

Now, we can find the sum of the roots \( S \):
\[ S = \alpha^2 + \beta^2 + \alpha^{-2} + \beta^{-2} = \frac{b^2 - 2ac}{a^2} + \frac{b^2 - 2ac}{c^2} \]
To combine these fractions, we need a common denominator:
\[ S = \frac{(b^2 - 2ac)c^2 + (b^2 - 2ac)a^2}{a^2c^2} = \frac{(b^2 - 2ac)(c^2 + a^2)}{a^2c^2} \]

### Step 4: Product of the Roots

Next, we find the product of the roots \( P \):
\[ P = (\alpha^2 + \beta^2)(\alpha^{-2} + \beta^{-2}) = \left(\frac{b^2 - 2ac}{a^2}\right)\left(\frac{b^2 - 2ac}{c^2}\right) = \frac{(b^2 - 2ac)^2}{a^2c^2} \]

### Step 5: Form the Quadratic Equation

Using the sum and product of the roots, we can form the quadratic equation:
\[ x^2 - Sx + P = 0 \]
Substituting \( S \) and \( P \):
\[ x^2 - \frac{(b^2 - 2ac)(c^2 + a^2)}{a^2c^2}x + \frac{(b^2 - 2ac)^2}{a^2c^2} = 0 \]
Multiplying through by \( a^2c^2 \) to eliminate the denominators gives:
\[ a^2c^2x^2 - (b^2 - 2ac)(c^2 + a^2)x + (b^2 - 2ac)^2 = 0 \]

### Final Answer

The quadratic equation whose roots are \( \alpha^2 + \beta^2 \) and \( \alpha^{-2} + \beta^{-2} \) is:
\[ a^2c^2x^2 - (b^2 - 2ac)(c^2 + a^2)x + (b^2 - 2ac)^2 = 0 \]