If α, β are roots of:

12x² − 20x + 3λ = 0,

where λ ∈ Z and

1 < |β − α| ≤ 3/2,

find sum of all possible values of λ.

Solution:

For quadratic equation:

ax² + bx + c = 0

Difference between roots:

|β − α| = √D / a

Discriminant:

D = (−20)² − 4(12)(3λ)

= 400 − 144λ

So,

|β − α| = √(400 − 144λ) / 12

Given:

1 < √(400 − 144λ)/12 ≤ 3/2

Multiply by 12:

12 < √(400 − 144λ) ≤ 18

Squaring:

144 < 400 − 144λ ≤ 324

Now solve:

144λ < 256

λ < 16/9

and

144λ ≥ 76

λ ≥ 19/36

Possible integer values:

λ = 1

Therefore sum = 1

Answer: 1