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Homework Statement:
 The equation 2x^2  2(2a + 1)x + a(a  1) = 0 has two real roots x1 and x2. Find a such that x1 < a < x2 !
Relevant Equations:

Discriminant
Quadratic formula
For quadratic equation to have two real roots:
b^{2}  4ac > 0
(2 (2a + 1))^{2}  4 (2) (a (a  1)) > 0
4 (4a^{2} + 4a + 1)  8a^{2} + 8a > 0
16a^{2} + 16a + 4  8a^{2} + 8a > 0
8a^{2} + 24 a + 4 > 0
2a^{2} + 6a + 1 > 0
Using quadratic formula, I get a < (3  √7) / 2 or a > (3 + √7) / 2
Then how to know if a is between x_{1} and x_{2}? Thanks
b^{2}  4ac > 0
(2 (2a + 1))^{2}  4 (2) (a (a  1)) > 0
4 (4a^{2} + 4a + 1)  8a^{2} + 8a > 0
16a^{2} + 16a + 4  8a^{2} + 8a > 0
8a^{2} + 24 a + 4 > 0
2a^{2} + 6a + 1 > 0
Using quadratic formula, I get a < (3  √7) / 2 or a > (3 + √7) / 2
Then how to know if a is between x_{1} and x_{2}? Thanks