\(\alpha\) and \(\beta\) are the roots of the equation \(2x^{2} - 3x + 4 = 0\). Find \(\frac{\alpha}{\beta} + \frac{\beta}{\alpha}\)

A.

\(\frac{-9}{8}\)

B.

\(\frac{-7}{8}\)

C.

\(\frac{7}{8}\)

D.

\(\frac{9}{8}\)

Correct answer is B

\(\frac{\alpha}{\beta} + \frac{\beta}{\alpha} = \frac{\alpha^{2} + \beta^{2}}{\alpha\beta}\)

\(\alpha + \beta = \frac{-b}{a}\); \(\alpha\beta = \frac{c}{a}\)

\(\alpha^{2} + \beta^{2} = (\alpha + \beta)^{2} - 2(\alpha\beta)\)

From the equation, a = 2, b = -3, c =4

\(\alpha + \beta = \frac{-(-3)}{2} = \frac{3}{2}\)

\(\alpha\beta = \frac{4}{2} = 2\)

\(\alpha^{2} + \beta^{2} = (\frac{3}{2})^{2} - 2(2)) = \frac{9}{4} - 4 = \frac{-7}{4}\)

\(\implies \frac{\alpha}{\beta} + \frac{\beta}{\alpha} = \frac{\frac{-7}{4}}{2} = \frac{-7}{8}\)