Final answer:
The equation x^2 - (2m-3)x + m^2 + m = 0 requires the discriminant to be positive for two distinct real solutions, leading to m < 9/16. For part b, we calculate |x1^2 - x2^2| using Vieta's formulas, while for part c, x1^2*x2 + x2^2*x1 is found in a similar manner, also using Vieta's formulas.
Step-by-step explanation:
The equation given is of the form x^2 - (2m-3)x + m^2 + m = 0, which is a quadratic equation where a = 1, b = -(2m-3), and c = m^2 + m. For the equation to have two distinct real solutions, the discriminant b^2 - 4ac must be positive.
a) To find the value of m for which the equation has two distinct solutions, we calculate the discriminant: D = b^2 - 4ac = (2m-3)^2 - 4(1)(m^2+m). This simplifies to D = 4m^2 - 12m + 9 - 4m^2 - 4m. Simplifying further, we get D = -16m + 9. Setting D > 0 gives us m < 9/16.
b) We know the identity |x1^2 - x2^2| = |(x1 + x2)(x1 - x2)|. The sum of the roots x1 + x2 is equal to -b/a and the product x1*x2 is c/a. Substituting b and c gives us |(-(-2m+3)/1)^2 - (m^2 + m)/1^2|.
c) For the third part, we use the identities derived from Vieta's formulas: x1^2*x2 + x2^2*x1 = (x1 + x2)(x1*x2). Again using the sum and product of roots (-b/a and c/a respectively), we get the expression ((-(-2m+3)/1)*(m^2 + m)/1).