Root it out

$\overline{xxyy} = 1100x+11y = 11(100x+y)$

To make $\overline{xxyy}$ perfect square, $100x+y$ shall have 11 as a factor, and both X and Y shall be within 1 to 9. Upon simplifying $100x+y$ to $11(9x)+(x+y)$ , we will also realize that $x+y = 11$ and then $9x+1$ shall be perfect square. Only $(x,y) = (7,4)$ satisfying this condition, which gives $88^2 = 7744$

AB^{2}=XXYY
AB= \sqrt{X 1100+YY 11})
AB= \sqrt{11}* \sqrt{X 100+Y}
Since AB is a Two-Digit number .Therefore, \sqrt{X
100+Y} should have one \sqrt{11} . It means AB is having 11 as its factor. Now the possible value of AB are 11,22,33,44,55,66,77,88,99 . when we square all of these we get our answer from squaring 88^{2} =7744