For which condition on \(s\) will the equation \(x^2-2sx+s+2=0\) have real and distinct roots?

Correct answer: \(s<-1\) या \(s>2\)

A quadratic equation has real and distinct roots only when its discriminant satisfies \(D>0\). Here, \(a=1\), \(b=-2s\), and \(c=s+2\), so \(D=(-2s)^2-4(1)(s+2)=4(s-2)(s+1)\). Thus, \((s-2)(s+1)>0\), which gives \(s<-1\) or \(s>2\). For \(-1<s<2\), \(D<0\), while \(s=-1\) or \(s=2\) gives \(D=0\). Exam tip: For real and distinct roots, always check that \(D>0\).