If the equation \\(3x^2+(k-2)x+4=0\\) has two distinct real roots, which condition on \\(k\\) is correct?

Correct answer: or \\(k>2+4\\sqrt{3}\\)

For a quadratic equation, the discriminant is \\(D=b^2-4ac\\). Here, \\(a=3\\), \\(b=k-2\\), and \\(c=4\\), so \\(D=(k-2)^2-48\\). Two distinct real roots require \\(D>0\\), which gives \\((k-2)^2>48\\). Hence \\(|k-2|>4\\sqrt{3}\\), so \\(k<2-4\\sqrt{3}\\) or \\(k>2+4\\sqrt{3}\\). Therefore, option A is correct. In option B, the discriminant is negative, and at the boundary values the roots are equal. Exam tip: for distinct real roots, always check \\(D>0\\).