Value of "a" between two real roots of quadratic equation

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SUMMARY

The discussion focuses on determining the values of "a" for which the quadratic equation 2x² - 2(2a + 1)x + a(a - 1) = 0 has two real roots, x1 and x2. The condition for two real roots is established through the discriminant, requiring that b² - 4ac > 0. The derived inequality simplifies to 2a² + 6a + 1 > 0, leading to the critical values a < (-3 - √7) / 2 or a > (-3 + √7) / 2. The participants discuss how to ascertain if "a" lies between the roots x1 and x2.

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songoku
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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:

b2 - 4ac > 0

(-2 (2a + 1))2 - 4 (2) (a (a - 1)) > 0

4 (4a2 + 4a + 1) - 8a2 + 8a > 0

16a2 + 16a + 4 - 8a2 + 8a > 0

8a2 + 24 a + 4 > 0

2a2 + 6a + 1 > 0

Using quadratic formula, I get a < (-3 - √7) / 2 or a > (-3 + √7) / 2Then how to know if a is between x1 and x2? Thanks
 
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songoku said:
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:

b2 - 4ac > 0

(-2 (2a + 1))2 - 4 (2) (a (a - 1)) > 0

4 (4a2 + 4a + 1) - 8a2 + 8a > 0

16a2 + 16a + 4 - 8a2 + 8a > 0

8a2 + 24 a + 4 > 0

2a2 + 6a + 1 > 0

Using quadratic formula, I get a < (-3 - √7) / 2 or a > (-3 + √7) / 2Then how to know if a is between x1 and x2? Thanks
Why not calculate ##x_1## and##x_2##?
 
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PeroK said:
Why not calculate ##x_1## and##x_2##?
I understand. Thank you very much PeroK
 

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