C-05-Q058mediumsingle_mcqAfter substituting p=axp = a^xp=ax, the equation a2x−(a2+1)ax+a2=0a^{2x} - (a^2+1)a^x + a^2 = 0a2x−(a2+1)ax+a2=0 becomes —ap2−(a2+1)p+a2=0p^2 - (a^2+1)p + a^2 = 0p2−(a2+1)p+a2=0bp2+(a2+1)p−a2=0p^2 + (a^2+1)p - a^2 = 0p2+(a2+1)p−a2=0cp2−a2p+1=0p^2 - a^2p + 1 = 0p2−a2p+1=0dp−(a2+1)+a2=0p - (a^2+1) + a^2 = 0p−(a2+1)+a2=0