What condition on \(k\) is necessary for the equation \(x^2+6x+k=0\) to have no real roots?

Correct answer: \(k>9\)

For this quadratic equation, \(a=1\), \(b=6\), and \(c=k\). Its discriminant is \(D=b^2-4ac=36-4k\). No real roots occur when \(D<0\), so \(36-4k<0\), which gives \(k>9\). At \(k=9\), the equation has one repeated real root, while \(k<9\) gives two real roots. Exam tip: Check the sign of the discriminant first to determine the nature of the roots.