If the equation \(x^2+tx+12=0\) has no real roots, what is the correct condition on \(t\)?

Correct answer: \(t^2<48\)

A quadratic equation \(ax^2+bx+c=0\) has no real roots when its discriminant \(D=b^2-4ac\) is negative. Here, \(a=1, b=t, c=12\), so \(D=t^2-48\). Thus, \(t^2-48<0\), which gives \(t^2<48\). If \(t^2=48\), the equation has two equal real roots, so option B is incorrect. Exam tip: determine the nature of the roots by checking the sign of the discriminant first.