To find the value of \( A \) for which the difference of the squares of the roots of the quadratic equation \( x^2 - (A - 3)x - (A - 2) = 0 \) is zero, we start by noting that the difference of the squares of the roots \( r_1 \) and \( r_2 \) can be expressed as: \[ r_1^2 - r_2^2 = (r_1 - r_2)(r_1 + r_2) \] For this difference to be zero, either \( r_1 - r_2 = 0 \) (which means the roots are equal) or \( r_1 + r_2 = 0 \). Using Vieta's formulas, we know: The sum of the roots \( r_1 + r_2 = A - 3 \) The product of the roots \( r_1 r_2 = -(A - 2) \) For the roots to be equal, the discriminant must be zero: \[ D = b^2 - 4ac = (A - 3)^2 - 4(1)(-(A - 2)) = 0 \] Expanding this gives: \[ (A - 3)^2 + 4(A - 2) = 0 \] Expanding further: \[ A^2 - 6A + 9 + 4A - 8 = 0 \] Combining like terms: \[ A^2 - 2A + 1 = 0 \] Factoring: \[ (A - 1)^2 = 0 \] Thus, \( A - 1 = 0 \) implies \( A = 1 \). Therefore, the value of \( A \) for which the difference of the squares of the roots is zero is: \[ \boxed{1} \]