If \(a\) and \(b\) are real numbers and the roots of \(x^2-2(a+b)x+3ab=0\) are real, which statement about \(a\) and \(b\) is correct?

Correct answer: The roots are real for all real \(a,b\) because \(a^2-ab+b^2\geq 0\)

The discriminant is \(D=[-2(a+b)]^2-4(1)(3ab)=4(a^2-ab+b^2)\). Since \(a^2-ab+b^2=\frac{1}{2}[(a-b)^2+a^2+b^2]\geq 0\), we have \(D\geq 0\), so the roots are real for every pair of real values of \(a\) and \(b\). Option B is not necessary; for example, when \(a=1,b=-1\), \(ab<0\) but the roots are still real. Equal roots require \(D=0\), which occurs only when \(a=b=0\), so option D is incorrect. Exam tip: For a quadratic equation, determine the nature of its roots by checking the sign of the discriminant.