Prove that $6\sqrt{3}$ is irrational.
The common difference of the A.P.: $3,\,3+\sqrt{2},\,3+2\sqrt{2},\,3+3\sqrt{2},\,\ldots$ will be:
The discriminant of the quadratic equation $3x^2 - 4\sqrt{3}\,x + 4 = 0$ will be: