Quadratic Equations

Given the quadratic equation x2+ax+b=0, we know from the problem statement:
The sum of the roots, α+β, is ‌

1

2

.
The sum of the cubes of the roots, α3+β3, is ‌

37

8

.
From Vieta's formulas:
α+β=−a=‌

1

2

. Thus, a=−‌

1

2

.
The product of the roots αβ=b.
We are given that α3+β3=‌

37

8

. Using the identity for the sum of cubes:
α3+β3=(α+β)3−3αβ(α+β)
Substituting the known values:
α3+β3=(‌

1

2

)3−3b(‌

1

2

)
Substitute the given value for α3+β3 :
‌

37

8

=‌

1

8

−‌

3b

2

Rearranging gives us:
‌

3b

2

=‌

1

8

−‌

37

8

=‌

−36

8

=‌

−9

2

Thus,
3b=−9‌‌⇒‌‌b=−3

To find a−‌

1

b

:
a−‌

1

b

=−‌

1

2

−‌

1

−3

=−‌

1

2

+‌

1

3

Calculate the above expression:
a−‌

1

b

=‌

−3

6

+‌

2

6

=‌

−1

6