Tolong di jawab kak .pakai cara ​

2x²-8x+12= 0

(2x-6)(x-2) = 0

2x-6 =0 , x-2 =0

2x =0+6 , x = 0+2

2x = 6 , x = 2

x =6÷2

x=3

jadi x1 = 3 dan x2 =2

a. x1+x2

= 3+2

= 5

b. x1.x2

= 3.2

= 6

c. x1²+x2²

= 3² + 2²

= 9 + 4

= 13

d. x1–x2

= 3-2

=1

2x^2 – 8x + 12 = 0

a = 2

b = -8

c = 12

a]

x1 + x2

= -b/a

= -(-8)/2

= 8/2

= 4

b]

x1 . x2

= c/a

= 12/2

= 6

c]

(x1)^2 + (x2)^2

= (x1 + x2)^2 – 2(x1 . x2)

= 4^2 – 2(6)

= 16 – 12

= 4

d]

x1 – x2

= (VD)/a

= [V(b^2 – 4ac)]/a

= [V((-8)^2 – 4(2)(12))]/2

= [V(64 – 96)]/2

= (V(-32))/2

= (1/2)V(-32)

(tidak terdefinisi karena akarnya bernilai negatif)