Jika x1 dan x2 akar-akar dari persamaan 4x^2+12x+5 = 0.

a. 13/2, b. -73/4, c. 0, d. -3/2

Pembahasan

Diberikan persamaan kuadrat 4x^2 + 12x + 5 = 0. Akar-akarnya adalah x1 dan x2.

Menurut teorema Vieta:
Jumlah akar (x1 + x2) = -b/a = -12/4 = -3
Perkalian akar (x1 * x2) = c/a = 5/4

a. Nilai dari x1^2 + x2^2
x1^2 + x2^2 = (x1 + x2)^2 - 2x1x2
x1^2 + x2^2 = (-3)^2 - 2(5/4)
x1^2 + x2^2 = 9 - 10/4
x1^2 + x2^2 = 9 - 5/2
x1^2 + x2^2 = 18/2 - 5/2
x1^2 + x2^2 = 13/2

b. Nilai dari x1^3 + x2^3 - 2x1x2
x1^3 + x2^3 = (x1 + x2)(x1^2 - x1x2 + x2^2)
x1^3 + x2^3 = (x1 + x2)((x1 + x2)^2 - 3x1x2)
x1^3 + x2^3 = (-3)((-3)^2 - 3(5/4))
x1^3 + x2^3 = (-3)(9 - 15/4)
x1^3 + x2^3 = (-3)(36/4 - 15/4)
x1^3 + x2^3 = (-3)(21/4)
x1^3 + x2^3 = -63/4

Maka, x1^3 + x2^3 - 2x1x2 = -63/4 - 2(5/4)
= -63/4 - 10/4
= -73/4

c. Nilai dari (2x1 + 5)(2x2 + 5)
(2x1 + 5)(2x2 + 5) = 4x1x2 + 10x1 + 10x2 + 25
= 4x1x2 + 10(x1 + x2) + 25
= 4(5/4) + 10(-3) + 25
= 5 - 30 + 25
= 0

d. Nilai dari 1/2x1 + 1/2x2
1/2x1 + 1/2x2 = 1/2 (x1 + x2)
= 1/2 (-3)
= -3/2

Jadi:
a. x1^2 + x2^2 = 13/2
b. x1^3 + x2^3 - 2x1x2 = -73/4
c. (2x1 + 5)(2x2 + 5) = 0
d. 1/2x1 + 1/2x2 = -3/2