Let P= x x^2+(n-1) x-2(n+1)=0 and Q= (n-1) x^2+n x+1=0 . Then find the number of values of n such that P Q has exactly 3 distinct elements (where x is a real number).

Step-by-step solution

Notice that x=2 is a root of P because 2^2+(n-1) 2-2(n+1)=0 and x=-1 is a root of Q because (n-1) (-1)^2+n (-1)+1=0 . Therefore, we see that P(x)=(x-2)(x+n+1) and Q(x)=(x+1)((n-1) x+1) If |P Q|=3 , then P(x) and Q(x) share a root. Therefore, we have the three equations: aligned 2 & =- ( 1 n-1 ) n= 1 2 -1=-n-1 n=0 - ( 1 n-1 ) & =-n-1 n= 2 aligned Hence, there are total of 7 possible values of n .