>50 ⇒x2−50x+100>0 Consider the equation x2−50x+100=0

D=b2−4ac=(−50)2−4×1×100 =2500−400=2100

⇒x=

50±√2100

2

=25±5√21 =2.08 or 47.9 ∴x2−50x+100>0 ⇒(x−2.08)(x−47.9)>0 ⇒x<2.08 or x>47.9 So,x=1,2,48,49,..,100 Thus, the number of values of x satisfying the given inequality are 55.