Jika akar-akar persamaan x^2+ax+b memenuhi persamaan 2x^2-(a+3)x+(3b-2)=0, maka...

x^2 + ax + b = 0
x1 + x2 = -a
x1x2 = b

2x^2-(a+3)x+(3b-2)= 0
x1 + x2 = -a-3/2
x1x2 = 3b-2/2

-a = -a-3/2
-2a = -a-3
-2a+a = -3
-a = -3
a = 3

b = 3b-2/2
2b = 3b-2
2b-3b = -2
-b = -2
b = 2

Jadi persamaannya adalah :
x^2+3x+2

jawab

(i) x² + ax + b = 0
(ii) 2x² -(a+3)x + (3b -2)= 0

2(i) = (ii)
2(x² + ax + b) = 2x² - (a+3)x + (3b -2)
2x² + 2ax + 2b= 2x² -(a+3)x + (3b -2)

2a = -(a+3)
2a = -a - 3
3a = - 3
a =  -1

2b = 3b -2
-b = - 2
b = 2

maka
a= - 1, b = 2