Question

Show that one zero of 8x230x27 is the square of the other.

Answer 1

8x2-30x+27 has one root (9/4) the square of the other (3/2)

Explanation:

8x2+30x+27 = (2*x+3)*(4*x+9) => solution = {-3/2, -4/9}

One is not the square of the other because squares are positive.

8x2-30x+27 = (2*x-3)*(4*x-9) =>

solution = {3/2, 4/9} since (4/9 = (3/2)^2, therefore the preceding question is correct.