\(\alpha\) and \(\beta\) are the roots of the equation \(2x^{2} - 3x + 4 = 0\). Find \(\frac{\alpha}{\beta} + \frac{\beta}{\alpha}\)
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Correct Answer: Option B
Explanation:
\(\frac{\alpha}{\beta} + \frac{\beta}{\alpha} = \frac{\alpha^{2} + \beta^{2}}{\alpha\beta}\)
\(\alpha + \beta = \frac{-b}{a}\); \(\alpha\beta = \frac{c}{a}\)
\(\alpha^{2} + \beta^{2} = (\alpha + \beta)^{2} - 2(\alpha\beta)\)
From the equation, a = 2, b = -3, c =4
\(\alpha + \beta = \frac{-(-3)}{2} = \frac{3}{2}\)
\(\alpha\beta = \frac{4}{2} = 2\)
\(\alpha^{2} + \beta^{2} = (\frac{3}{2})^{2} - 2(2)) = \frac{9}{4} - 4 = \frac{-7}{4}\)
\(\implies \frac{\alpha}{\beta} + \frac{\beta}{\alpha} = \frac{\frac{-7}{4}}{2} = \frac{-7}{8}\)
\(\frac{\alpha}{\beta} + \frac{\beta}{\alpha} = \frac{\alpha^{2} + \beta^{2}}{\alpha\beta}\)
\(\alpha + \beta = \frac{-b}{a}\); \(\alpha\beta = \frac{c}{a}\)
\(\alpha^{2} + \beta^{2} = (\alpha + \beta)^{2} - 2(\alpha\beta)\)
From the equation, a = 2, b = -3, c =4
\(\alpha + \beta = \frac{-(-3)}{2} = \frac{3}{2}\)
\(\alpha\beta = \frac{4}{2} = 2\)
\(\alpha^{2} + \beta^{2} = (\frac{3}{2})^{2} - 2(2)) = \frac{9}{4} - 4 = \frac{-7}{4}\)
\(\implies \frac{\alpha}{\beta} + \frac{\beta}{\alpha} = \frac{\frac{-7}{4}}{2} = \frac{-7}{8}\)