Answer
The proposed negation is wrong.
Correct Negation:
There exist real numbers $x_1$ and $x_2$ such that $x_1^2 = x_2^2$ and $x_1 \ne x_2 $
Work Step by Step
The proposed negation is wrong.
The negation of a “for all” statement is not a “for all” statement
The negation of an if-then statement is not an if-then statement.
Negation of the statements of the form, "For all x, q", q being a statement, is "There exists x, ~q".
In this case,
q : $ x_1 = x_2 .$
So, ~q: $x_1 \ne x_2 $.
