Let \(u\) and \(v\) be the solutions to \(3x^2+5x+7=x^2+4x\). Find \(\frac{u}{v}+\frac{v}{u}\).
We can apply Vieta's formula to this problem using the fact that \(\frac{u}{v}+\frac{v}{u}=\frac{u^2+v^2}{vu}=\frac{u^2+v^2+2uv-2uv}{uv}=\frac{(u+v)^2-2uv}{uv}\).
By moving all the terms to the left side, the equation transforms into \(2x^2+x+7=0\). So then, by Vieta's formula, \(u+v=-\frac{1}{2}\) and \(uv = \frac{7}{2}\).
Plugging this in to what we got, we have \(\frac{u}{v}+\frac{v}{u}=\frac{(-\frac{1}{2})^2-2\cdot\frac{7}{2}}{\frac{7}{2}}=-\frac{27}{14}\).