For which values of \(a\) will the equation \(3x^2-2(2a+1)x+(a^2+a+1)=0\) have real and distinct roots?

Correct answer: \(a<-2\) or \(a>1\)

A quadratic equation has real and distinct roots when its discriminant satisfies \(D>0\). Here, \(A=3\), \(B=-2(2a+1)\), and \(C=a^2+a+1\). Thus, \(D=B^2-4AC=4(a^2+a-2)=4(a+2)(a-1)\). Therefore, \((a+2)(a-1)>0\), which gives \(a<-2\) or \(a>1\). At \(a=-2\) or \(a=1\), \(D=0\), so the roots are equal rather than distinct. Exam tip: \(D>0\), \(D=0\), and \(D<0\) indicate real distinct, real equal, and non-real roots, respectively.