Question

Bagaimana cara mengoperasikan bentuk akar kubik ini?
[tex]\displaystyle \sqrt[3]{9+4\sqrt{5}}+\sqrt[3]{9-4\sqrt{5}}[/tex]

Answer 1

Verified answer

Jawab:
= 3

Penjelasan:
Pakai permisalan
[tex]\displaystyle p=\sqrt[3]{9+4\sqrt{5}} ,\:\:\:q=\sqrt[3]{9-4\sqrt{5}} \\p^3=9+4\sqrt{5},\:\:\:\:\:\:q^3=9-4\sqrt{5}[/tex]
Karena kesekawanan
[tex]\displaystyle p^3=\frac{1}{q^3},\:\:\:q^3=\frac{1}{p^3},\:\:\:p^3q^3=1\\~~~~~~~~~~~~~~~~~~~~~~~~~~~\:(pq)^3=1\\~~~~~~~~~~~~~~~~~~~~~~~~~~~\:\:\:\:\:\:\:pq=1[/tex]
Pakai permisalan lagi
p + q = x
x³ = (p+q)³
x³ = p³ + 3p²q + 3pq² + q³
x³ = p³ + 3p(pq) + 3q(pq) + q³
x³ = p³ + 3p(1) + 3q(1) + q³
x³ = p³ + 3(p + q) + q³
Maka
[tex]\displaystyle x^3=9+4\sqrt{5}+3(p+q)+9-4\sqrt{5}\\x^3=9+3(p+q)+9\\x^3=3(p+q)+18\\\because p+q=x\therefore\\x^3=3x+18\\x^3-3x-18=0[/tex]
Selesaikan
[tex]\displaystyle x^3+3x^2-3x^2+6x-9x-18=0\\x^3+3x^2+6x-3x^2-9x-18=0\\x(x^2+3x+6)-3(x^2+3x+6) = 0\\(x-3)(x^2+3x+6) =0[/tex]
Untuk x-3 = 0, x = 3
Untuk x² - 3x + 6 = 0, cek diskriminan (d)
d = b² - 4ac
d = (-3)² - 4(1)(6)
d = 9 - 24 (solusi tidak real krn d = negatif)
x = 3
maka p + q = 3

[tex]\sqrt[3]{9+4\sqrt{5}} +\sqrt[3]{9-4\sqrt{5}} =\bf3[/tex]
(xcvi)