Browse · harp Print → smc algebra senior Problem The roots of the equation 2x+2x−21=5 can be found by solving: (A) 16x2−92x+1=0 (B) 4x2−25x+4=0 (C) 4x2−17x+4=0 (D) 2x2−21x+2=0 Solution — click to reveal Make the substitution t=x. Then 2t+t2=5⟹2t2−5t+2=0. Then 2x−5x+2=0⟹2x+2=5x⟹(2x+2)2=25x⟹4x2+8x+4=25x⟹4x2−17x+4=0, 4x2−17x+4=0 Final answer C ← Previous problem Next problem →