The equation \(x^2-2(4t+1)x+(7t^2+2t+5)=0\) has no real roots. Which is the correct interval for \(t\)?

Correct answer: \(\frac{-1-\sqrt{5}}{3}<t<\frac{-1+\sqrt{5}}{3}\)

For a quadratic equation, real roots do not exist when the discriminant \(D<0\). Here, \(D=[-2(4t+1)]^2-4(7t^2+2t+5)=4(9t^2+6t-4)\). Thus, \(9t^2+6t-4<0\). Its zeros are \(\frac{-1-\sqrt5}{3}\) and \(\frac{-1+\sqrt5}{3}\), and since the quadratic has a positive leading coefficient, the expression is negative between these zeros. Hence option A is correct. At the endpoints, \(D=0\), giving equal real roots, so the endpoints must be excluded. Exam tip: for no real roots, always use the strict condition \(D<0\), not \(D\leq0\).