If `alpha , beta ` are the roots of the equation ` x^(2)-px+q=0`, then ` (alpha^(2))/(beta^(2))+(beta^(2))/(alpha^(2))` is equal to :

Options

A. `(p^4)/(q^(2))+2-(4p^(2))/(q)`
B. `(p^4)/(q^(2))-2+(4p^(2))/(q)`
C. `(p^(4))/(q^(2))+2q^(2)-(4p^(2))/(q)`
D. None of these

Answer

To solve the problem, we start with the quadratic equation given:

\[ x^2 - px + q = 0 \]

Let \(\alpha\) and \(\beta\) be the roots of this equation. According to Vieta's formulas, we know:

1. The sum of the roots:
\[ \alpha + \beta = p \]

2. The product of the roots:
\[ \alpha \beta = q \]

We need to find the value of the expression:

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

This expression can be rewritten as:

\[ \frac{\alpha^4 + \beta^4}{\alpha^2 \beta^2} \]

Next, we need to find \(\alpha^4 + \beta^4\). We can use the identity:

\[ \alpha^4 + \beta^4 = (\alpha^2 + \beta^2)^2 - 2\alpha^2\beta^2 \]

To use this identity, we first need to find \(\alpha^2 + \beta^2\). We know:

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

Substituting the values from Vieta's formulas:

\[ \alpha^2 + \beta^2 = p^2 - 2q \]

Now, substituting \(\alpha^2 + \beta^2\) into the identity for \(\alpha^4 + \beta^4\):

\[ \alpha^4 + \beta^4 = (p^2 - 2q)^2 - 2q^2 \]

Now we expand \((p^2 - 2q)^2\):

\[ (p^2 - 2q)^2 = p^4 - 4pq + 4q^2 \]

Thus,

\[ \alpha^4 + \beta^4 = p^4 - 4pq + 4q^2 - 2q^2 = p^4 - 4pq + 2q^2 \]

Now we substitute this back into our expression for \(\frac{\alpha^2}{\beta^2} + \frac{\beta^2}{\alpha^2}\):

\[ \frac{\alpha^4 + \beta^4}{\alpha^2 \beta^2} = \frac{p^4 - 4pq + 2q^2}{q^2} \]

This can be simplified as:

\[ \frac{p^4}{q^2} - \frac{4p}{q} + 2 \]

Thus, the final answer is:

\[ \frac{p^4}{q^2} - \frac{4p}{q} + 2 \]