If \(\alpha\) and \(\beta\) are the roots of the equation \(2x^{2} - 7x + 4 = 0\), find the equation whose roots are \(\frac{\alpha}{\beta}\) and \(\frac{\beta}{\alpha}\).
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Correct Answer: Option n
Explanation:
\(2x^{2} - 7x + 4 = 0\)
\(\alpha + \beta = \frac{7}{2}\)
\(\alpha \beta = \frac{4}{2} = 2\)
\(\frac{\alpha}{\beta} + \frac{\beta}{\alpha} = \frac{\alpha^{2} + \beta^{2}}{\alpha \beta}\)
= \(\frac{(\alpha + \beta)^{2} - 2\alpha \beta}{\alpha \beta}\)
= \(\frac{(\frac{7}{2})^{2} - 2(2)}{2}\)
= \(\frac{(\frac{49}{4} - \frac{16}{4})}{2}\)
= \(\frac{33}{8}\)
\(\frac{\alpha}{\beta} \times \frac{\beta}{\alpha} = 1\)
\(\therefore Equation : x^{2} - \frac{33}{8}x + 1 = 0\)
\(2x^{2} - 7x + 4 = 0\)
\(\alpha + \beta = \frac{7}{2}\)
\(\alpha \beta = \frac{4}{2} = 2\)
\(\frac{\alpha}{\beta} + \frac{\beta}{\alpha} = \frac{\alpha^{2} + \beta^{2}}{\alpha \beta}\)
= \(\frac{(\alpha + \beta)^{2} - 2\alpha \beta}{\alpha \beta}\)
= \(\frac{(\frac{7}{2})^{2} - 2(2)}{2}\)
= \(\frac{(\frac{49}{4} - \frac{16}{4})}{2}\)
= \(\frac{33}{8}\)
\(\frac{\alpha}{\beta} \times \frac{\beta}{\alpha} = 1\)
\(\therefore Equation : x^{2} - \frac{33}{8}x + 1 = 0\)