Find the number of real number solutions for the equation. x^2-4x+4=0 a-1 b-0 c-cannot be determined d-2

X² - 4x + 4 = 0
(x-2)(x-2) = 0
the equation has one root mainly 2 with multiplicity 2
the answer is a

The first one, i believe, would be correct. all the others have one thing in common, and that's that they have repeating numbers. in the second one, x and y have repeating 6's and 3's, and in the third, x and y have repeating 6's as well. in the last one, there are repeating 3's. the first one has no repeating numbers.