Difficulty in Quadratic equation and realtion with irrational roots

Question

One root of the quadratic equation $ax^2 +bx + c=0$ is $\dfrac{2}{\sqrt{3} + \sqrt{5}}$. If $\frac{c}{a}$ is rational, then how do we find the other root. the answer given is that the other root is $q(\sqrt{5}-\sqrt{3})$. I don't understand why should the other root be a multiple of $q$. Kindly answer

Answer 1

Try writing this as $ax^2+bx+c=a(x-r_1)(x-r_2)$ where $r_1,r_2$ are your roots then expand and compare.

Answer 2

Hint: since $\,(\sqrt5 -\sqrt 3)(\sqrt 5 + \sqrt3) = 5\!-\!3 = 2,\,$ the first root is $\,2/(\sqrt5 + \sqrt3) = \sqrt5-\sqrt 3.\,$ Thus, by Vieta, if the constant term is the rational $\,q = c/a,\,$ then the other root is $\, q/(\sqrt5 - \sqrt 3).$ Hence it appears that there is a typo in the answer (the slash in the fraction was omitted).