Akar akar persamaan x^-5x+4=0,persamaan kuadrat baru dan akar akarnya 2P+1 dan 2Q+1

X²- 5x + 4= 0 akrnya p dan q
akar baru 2p+1 dan 2q + 1
x = 2p + 1
2p = x- 1
p = 1/2 x – 1/2
p = 1/2( x -1)
PK baru  , subsitusikan  1/2 (x-1) .
{1/2 (x-1)}² – 5 { 1/2 (x-1)} + 4=  0
1/4 (x-1)² – 5/2 ( x-1) + 4  = 0 ….kalikan 4
(x-1)² – 10(x-1) + 16 = 0
x² -2x +1 – 10x +10 +16 = 0
x² -12x  + 27= 0

PKB
——-

a=1 b=-5 c=4
(2p+1)+(2q+1)=-b/a=-(-5)/1=5
(2p+1)(2q+1)=c/a=4/1=4

*(2p+1)+(2q+1)
=2p+2q+2
=2(p+q)+2
=2(5)+2
=10+2
=12

*(2p+1)(2q+1)
=4pq+2p+2q+1
=4pq+2(p+q)+1
=4(4)+2(5)+1
=16+10+1
=27

PKB:
x^2-12x+27=0