The roots of the quadratic equation \(2x^2+kx+3=0\), where \(k>0\), are \(\alpha\) and \(\beta\), whilst those of the equation \(3x^2-2x+3\) are \(\dfrac{\alpha}{\beta}\) and \(\dfrac{\beta}{\alpha}\). Calculate the value of \(k\).

Since we know the roots of \(2x^2+kx+3=0\), we can write this as \[2(x-\alpha)(x-\beta)=0.\]

After expanding we can compare the coefficients with our original quadratic equation:\[2x^2+kx+3=2x^2-2(\alpha+\beta)x+2\alpha\beta.\] We then find

\[\begin{align} \label{eq:1} k&=-2(\alpha+\beta),\\ \label{eq:2}3&=2\alpha\beta. \end{align}\]

We can now do the same with the other quadratic, to obtain \(3x^2-2x+3=3\left(x-\dfrac{\alpha}{\beta}\right)\left(x-\dfrac{\beta}{\alpha}\right)=3x^2-3\left(\dfrac{\alpha}{\beta}+\dfrac{\beta}{\alpha}\right)x+3\) and hence

\[\begin{align*} -2&=-3\left(\dfrac{\alpha}{\beta}+\dfrac{\beta}{\alpha}\right)\\ &\implies \dfrac{2}{3} = \dfrac{\alpha^2+\beta^2}{\alpha\beta} \\ &\implies \dfrac{2}{3}\alpha\beta = (\alpha+\beta)^2-2\alpha\beta \\ &\implies (\alpha+\beta)^2 = 4\\ &\implies \dfrac{k^2}{4} = 4\\ &\implies k = 4. \end{align*}\]

We might note that if we substitute \(k=4\) into the original equation, and try to solve it, we find that \(\alpha\) and \(\beta\) are not real, but complex.