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kx 2 − (1 + k)x + (3k + 2) = 0 ()

.k

β α .c = 3k + 2 b = −(1 + k) a = k
k = −3
.
5
α + β = 2αβ

−b = 2ac 
a

1+k = 2  3k + 2 
k k

6k2 + 4k = k + k 2

5k2 + 3k = 0

k(5k + 3) = 0

( ) k =/ 0 .k = −3 k =0
k
5

.

−3x2 − 2x + 1 = 0
5 55

3x 2 + 2x − 1 = 0

(3x − 1)(x + 1) = 0

.x2 = −1 x1 = 1
3

. x 2 − 6x + 7 = 0 β α ()

.β + 1 α + 1

αβ