If ` alpha and beta ` are the roots of equation ` x^(2) - 2 x + 4 = 0 ` ,then what is the equation whose roots are `( alpha^(3))/( beta^(2)) and ( beta^(3))/( alpha^(2))` ?

Options

A. ` x ^(2) - 4 x + 8 =0`
B. ` x^(2) - 32 x + 4 =0`
C. ` x^(2) - 2x + 4 = 0`
D. `x^(2) - 16 x + 4 = 0 `

Answer

To find the equation whose roots are \( \frac{\alpha^3}{\beta^2} \) and \( \frac{\beta^3}{\alpha^2} \), where \( \alpha \) and \( \beta \) are the roots of the equation \( x^2 - 2x + 4 = 0 \), we will follow these steps:

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

The roots of the quadratic equation \( x^2 - 2x + 4 = 0 \) can be found using the quadratic formula:

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

Here, \( a = 1 \), \( b = -2 \), and \( c = 4 \).

Calculating the discriminant:

\[ b^2 - 4ac = (-2)^2 - 4 \cdot 1 \cdot 4 = 4 - 16 = -12 \]

Since the discriminant is negative, the roots are complex:

\[ x = \frac{2 \pm \sqrt{-12}}{2} = 1 \pm i\sqrt{3} \]

Thus, we have:

\[ \alpha = 1 + i\sqrt{3}, \quad \beta = 1 - i\sqrt{3} \]

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

First, we need to compute \( \alpha^3 \) and \( \beta^3 \).

Using the identity \( \alpha^2 = 2\alpha - 4 \):

\[ \alpha^2 = (1 + i\sqrt{3})^2 = 1 + 2i\sqrt{3} - 3 = -2 + 2i\sqrt{3} \]

Now, calculate \( \alpha^3 \):

\[ \alpha^3 = \alpha \cdot \alpha^2 = (1 + i\sqrt{3})(-2 + 2i\sqrt{3}) = -2 - 2i\sqrt{3} + 2i\sqrt{3} - 6 = -8 \]

Similarly, for \( \beta \):

\[ \beta^2 = (1 - i\sqrt{3})^2 = 1 - 2i\sqrt{3} - 3 = -2 - 2i\sqrt{3} \]

Calculating \( \beta^3 \):

\[ \beta^3 = \beta \cdot \beta^2 = (1 - i\sqrt{3})(-2 - 2i\sqrt{3}) = -2 + 2i\sqrt{3} + 2i\sqrt{3} - 6 = -8 \]

Now we can find:

\[ \frac{\alpha^3}{\beta^2} = \frac{-8}{-2 - 2i\sqrt{3}} = \frac{-8}{-2(1 + i\sqrt{3})} = \frac{4}{1 + i\sqrt{3}} \]

To simplify, multiply the numerator and denominator by the conjugate:

\[ \frac{4(1 - i\sqrt{3})}{1^2 + (i\sqrt{3})^2} = \frac{4(1 - i\sqrt{3})}{1 + 3} = \frac{4(1 - i\sqrt{3})}{4} = 1 - i\sqrt{3} \]

Similarly:

\[ \frac{\beta^3}{\alpha^2} = \frac{-8}{-2 + 2i\sqrt{3}} = \frac{4}{1 - i\sqrt{3}} = 1 + i\sqrt{3} \]

### Step 3: Form the new equation

The roots of the new equation are \( 1 - i\sqrt{3} \) and \( 1 + i\sqrt{3} \). The sum and product of the roots can be calculated as follows:

Sum of the roots:

\[ (1 - i\sqrt{3}) + (1 + i\sqrt{3}) = 2 \]

Product of the roots:

\[ (1 - i\sqrt{3})(1 + i\sqrt{3}) = 1^2 + 3 = 4 \]

Now, using the sum and product of the roots, we can write the new quadratic equation:

\[ x^2 - (\text{sum of roots})x + \text{product of roots} = 0 \]

Thus, the equation is:

\[ x^2 - 2x + 4 = 0 \]

### Final Answer

The equation whose roots are \( \frac{\alpha^3}{\beta^2} \) and \( \frac{\beta^3}{\alpha^2} \) is:

\[ x^2 - 2x + 4 = 0 \]