For the quadratic equation \(x^2-2kx+9=0\) to have real roots, which of the following conditions on \(k\) is required?

Correct answer: \(k\leq -3\) or \(k\geq 3\)

A quadratic equation \(ax^2+bx+c=0\) has real roots when its discriminant \(D=b^2-4ac\) is non-negative. Here, \(a=1, b=-2k, c=9\), so \(D=(-2k)^2-4(1)(9)=4k^2-36\). Thus, \(4k^2-36\geq0\), which gives \(k^2\geq9\), or \(k\leq-3\) or \(k\geq3\). Option B makes the discriminant negative, while option D is incomplete because \(k=-3\) also gives real equal roots. Exam tip: For real roots, always apply the condition \(D\geq0\).