A number problem leads to the equation \(n^2-2pn+(p^2-11p)=0\). What condition on \(p\) is necessary for the equation to have two real and distinct values of \(n\)?

Correct answer: \(p>0\)

For a quadratic equation, the discriminant is \(D=b^2-4ac\). Here, \(a=1\), \(b=-2p\), and \(c=p^2-11p\), so \(D=(-2p)^2-4(p^2-11p)=44p\). Two real and distinct roots require \(D>0\); hence \(44p>0\), which gives \(p>0\). Note that when \(p=0\), \(D=0\), so the roots are equal rather than distinct.