Explanation

For the equation to be an identity in x, each coefficient must vanish separately. From the x² coefficient, k²−5k+4=0 gives k=1 or k=4; from the x coefficient, k²−3k−4=0 gives k=4 or k=−1; from the constant term, k²−4k=0 gives k=0 or k=4. The only value satisfying all three conditions simultaneously is k=4, option D. Checking only the x² coefficient wrongly admits k=1 as well.