If `alpha,beta` are the roots of `ax^(2)+bx+c=0`, form that equation whose roots are `(alpha^(2)+beta^(2)),((1)/(alpha^(2))+(1)/(beta^(2)))`.

Answer

To find the quadratic equation whose roots are \( \alpha^2 + \beta^2 \) and \( \frac{1}{\alpha^2} + \frac{1}{\beta^2} \), we will follow these steps:

### Step 1: Find \( \alpha + \beta \) and \( \alpha \beta \)

From the given quadratic equation \( ax^2 + bx + c = 0 \), we know:

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

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

Using the identity:

\[ \alpha^2 + \beta^2 = (\alpha + \beta)^2 - 2\alpha\beta \]

Substituting the values we found in Step 1:

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

Calculating this gives:

\[ \alpha^2 + \beta^2 = \frac{b^2}{a^2} - \frac{2c}{a} = \frac{b^2 - 2ac}{a^2} \]

### Step 3: Calculate \( \frac{1}{\alpha^2} + \frac{1}{\beta^2} \)

Using the identity:

\[ \frac{1}{\alpha^2} + \frac{1}{\beta^2} = \frac{\beta^2 + \alpha^2}{\alpha^2 \beta^2} \]

Substituting the values we have:

\[ \frac{1}{\alpha^2} + \frac{1}{\beta^2} = \frac{\alpha^2 + \beta^2}{\alpha^2 \beta^2} \]

We already calculated \( \alpha^2 + \beta^2 \) in Step 2. 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,

\[ \frac{1}{\alpha^2} + \frac{1}{\beta^2} = \frac{\frac{b^2 - 2ac}{a^2}}{\frac{c^2}{a^2}} = \frac{b^2 - 2ac}{c^2} \]

### Step 4: Find the sum of the new roots

Now we can find the sum of the new roots:

\[ \text{Sum} = \left(\alpha^2 + \beta^2\right) + \left(\frac{1}{\alpha^2} + \frac{1}{\beta^2}\right) = \frac{b^2 - 2ac}{a^2} + \frac{b^2 - 2ac}{c^2} \]

Finding a common denominator:

\[ \text{Sum} = \frac{(b^2 - 2ac)c^2 + (b^2 - 2ac)a^2}{a^2c^2} = \frac{(b^2 - 2ac)(a^2 + c^2)}{a^2c^2} \]

### Step 5: Find the product of the new roots

Next, we find the product of the new roots:

\[ \text{Product} = \left(\alpha^2 + \beta^2\right) \left(\frac{1}{\alpha^2} + \frac{1}{\beta^2}\right) = \left(\frac{b^2 - 2ac}{a^2}\right) \left(\frac{b^2 - 2ac}{c^2}\right) = \frac{(b^2 - 2ac)^2}{a^2c^2} \]

### Step 6: Form the new quadratic equation

The new quadratic equation can be formed using the sum and product of the roots:

\[ x^2 - \text{(Sum)} \cdot x + \text{(Product)} = 0 \]

Substituting the values we found:

\[ x^2 - \frac{(b^2 - 2ac)(a^2 + c^2)}{a^2c^2} x + \frac{(b^2 - 2ac)^2}{a^2c^2} = 0 \]

### Final Form

Multiplying through by \( a^2c^2 \) to eliminate the denominators gives:

\[ a^2c^2x^2 - (b^2 - 2ac)(a^2 + c^2)x + (b^2 - 2ac)^2 = 0 \]

This is the required quadratic equation whose roots are \( \alpha^2 + \beta^2 \) and \( \frac{1}{\alpha^2} + \frac{1}{\beta^2} \).

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