persamaan kuadrat ax^2 + x + b = 0 mempunyai akar akar x1 dan x2.jika persamaan kuadrat ax^2 + 7x + 5 = 0 mempunyai akar akar x1 - 3 dan x1 - 3 maka nilai a.b =

jawab

ax² + x + b = 0..(i)
Pk baru yg akarnya x1 -3 dan  x2 -3
misal
p = x1 - 3
x1= p+ 3
sub ke ..(i)
a(p+3)² + (p+3) + b = 0
a (p² + 6p+ 9) +p + 3 + b = 0
ap² +6ap +9a + p+ 3 + b = 0
ubah p= x
ax² +6a x + 9a + x + 3 + b = 0
ax² +(6a +1) x + (9a + 3 +b) = ax² +7x + 5 = 0
6a +1 = 7
a = 1
9a + 3 + b = 5
9(1) + 3 +b= 5
12 + b = 5
b = - 7
a.b = - 7
.....
(cara yang lain)
......
ax² + x + b  0
x1+x2 = -1/a
x1. x2 = b/a
akar baru  p dan q
p =x1 - 3
q = x2 - 3
P+q = x1+x2 - 6 = (-1/a -6)
p q = (x1-3)(x2- 3) = (x1 x2 ) - 3(x1 + x2) + 9
p q = (b/a) - 3(- 1/a) + 9
p q = b/a + 3/a + 9

PK baru  x² -(p+q) x + (pq) =0
x² - (-1/a  - 6) x + (b/a + 3/a+9) =0
x² + 1/a x+ 6 x+ b/a + 3/a + 9 = 0....kalikan a
ax² + x + 6a x+ b + 3 + 9a = 0
ax² +(1 +6a) x + (b+3+9a) = 0
ax² + (1+6a) x + (b + 3+ 9a) = ax² +7x + 5
koefisien kiri = koefisien kanan
a= a
1+6a = 7 → a= 1
b+3 +9a = 5
b + 3 +9(1)= 5
b + 12 = 5
b = - 7

a. b = (1)(-7) = - 7