p(x)=0 ⇒f(x)=g(x) ⇒ax2+bx+c=a1x2+b1x+c1 ⇒(a−a1)x2+(b−b1)x+(c−c1)=0 It has only one solution, x=−1 ⇒b−b1=a−a1+c−c1...(i) Sum of roots
−(b−b1)
(a−a1)
=−1−1 ⇒
b−b1
2(a−a1)
=1 ........(ii) ⇒b−b1=2(a−a1) Now p(−2)=2 ⇒f(−2)−g(−2)=2 ⇒4a−2b+c−4a1+2b1−c1=2 ⇒4(a−a1)−2(b−b1)+(c−c1)=2...(iii) From equations, (i), (ii) and (iii) a−a1=c−c1=