lf c≠0 and the equation p(2x)=a(x+c)+b(x−c) has two equal roots, then p can be

We can write the given equation as p2x=(a+b)x+c(b−a)x2−c2or   p(x2−c2)=2(a+b)x2−2c(a−b)xor   (2a+2b−p)x2−2c(a−b)x+pc2=0For this equation to have equal roots,  c2(a−b)2−pc2(2a+2b−p)=0or  (a−b)2−2p(a+b)+p2=0or  [p−(a+b)]2=(a+b)2−(a−b)2=4abor  p−(a+b)=±2abor  p=a+b±2ab=(a±b)2 [∵c2≠0]