I have to prove that numbers $\displaystyle {\sqrt2}$ and $\displaystyle {\sqrt3}$ can't be solutions of quadratic equation if coefficients are rational.

My solution is based on using Viete's formulas.

$\displaystyle {\sqrt2}+{\sqrt3} = -\frac{b}{a}$

$\displaystyle {\sqrt2}{\sqrt3} = \frac{c}{a} $

From $\displaystyle {\sqrt2}{\sqrt3} = \frac{c}{a} $ we get that

$\displaystyle {a} = \frac{c}{{\sqrt2}{\sqrt3}} $.

So $\displaystyle {a} $ is then irrational number.

Knowing that, then $\displaystyle {b} $ is irrational number because

$\displaystyle -{b} = {a}({\sqrt2}+{\sqrt3}) =\frac{{c}({\sqrt2}+{\sqrt3})}{{\sqrt2}{\sqrt3}} $

c is then easy to prove that its also irrational.

Is my proof correct?