Akar akar persamaan kuadrat (p-2)x2+4x+(p+2) adalah a dan b . Jika a2b+ab2=-20maka p

A = p-2
B = 4
C = p+2

a+b = -B/A = -4/p-2
axb = C/A = p+2/p-2

a^2b + ab^2 = -20
(axb)(a + b) = -20
(p+2/p-2) (-4/p-2) = -20
-4(p+2)/(p-2)(p-2) = -20
-4p -8/p^2 -2p -2p + 4 = -20
-4p - 8/ p^2 -4p + 4 = -20
-4p -8 = -20(p^2 - 4p + 4)
-4p - 8 = -20p^2 +80p -80
20p^2 -80p +80 -4p -8 = 0
20p^2 -84p +72 = 0 (bagi 4)
5p^2 - 21p + 18= 0
(5p - 6)(p-3)
p = 6/5 atau p = 3

Mapel : Matematika
Kelas : VIII SMP
Bab : Persamaan Kuadrat

Pembahasan :
(p - 2)x² + 4x + (p + 2) = 0

Berdasarkan Teorema Vietta
X1 + X2 = -b/a
X1 + X3 = -4/(p - 2)

X1 . X2 = c/a
X1 . X2 = (p + 2)/(p - 2)

Maka...
a² + b² = -2
(a + b)² - 2ab = -2
[-4/(p - 2)]² - 2[(p + 2)/(p - 2)] = -2
16/(p - 2)² - (2p + 4)/(p - 2) = -2
[16 - (2p + 4)(p - 2)]/(p - 2)² = -2
[16 - (2p² - 8)] = -2(p - 2)²
-2p² + 24 = -2(p² - 4p + 4)
-2p² + 24 = -2p² + 8p - 8
8p = 32
P = 4