For the quadratic equation \(x^2-2px+(p^2+36)=0\), what is the correct conclusion about the nature of its roots for any real value of \(p\)?

Correct answer: No real roots

Here, \(a=1\), \(b=-2p\), and \(c=p^2+36\). Therefore, the discriminant is \(D=b^2-4ac=(-2p)^2-4(p^2+36)=-144\), which remains negative for every real value of \(p\). Hence, the equation has no real roots. Equal real roots would require \(D=0\), whereas here \(D<0\). Exam tip: For a quadratic equation, \(D<0\) immediately indicates that there are no real roots.