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{\b...

Answer Details

For \(2x^2 - 7x + 4 = 0\), the sum and product of the roots \(\alpha,\beta\) are

\[\alpha + \beta = \frac{7}{2},\qquad \alpha\beta = \frac{4}{2} = 2.\]

Sum of the new roots \(\dfrac{\alpha}{\beta} + \dfrac{\beta}{\alpha} = \dfrac{\alpha^2 + \beta^2}{\alpha\beta}\). Using \(\alpha^2 + \beta^2 = (\alpha+\beta)^2 - 2\alpha\beta\):

\[\alpha^2 + \beta^2 = \left(\frac{7}{2}\right)^2 - 2(2) = \frac{49}{4} - 4 = \frac{33}{4}.\]

\[\frac{\alpha}{\beta} + \frac{\beta}{\alpha} = \frac{33/4}{2} = \frac{33}{8}.\]

Product of the new roots \(\dfrac{\alpha}{\beta}\cdot\dfrac{\beta}{\alpha} = 1\).

The required equation is \(x^2 - (\text{sum})x + (\text{product}) = 0\):

\[x^2 - \frac{33}{8}x + 1 = 0 \;\Rightarrow\; 8x^2 - 33x + 8 = 0.\]