Jawab:
x = -3,-1,-4
Penjelasan dengan langkah-langkah:
Catatan : kalau latexnya error harap liat jawabannya di web
Apakah persamaanya x³-13x - 12 ?
[tex]f(x) = x^3-13x-12 = 0\\x^3 = 12+13x \to x = a+b \to a^3+b^3 = 12, 3ab = 13\to b = \dfrac{13}{3a}[/tex]
[tex]a^3+\left( \dfrac{13}{3a} \right)^3 = 12\to 27a^6 - 324a^3 + 2197 = 0\\\\a^3 = \dfrac{324 + \sqrt{324^2-4\cdot 27\cdot 2197}}{2\cdot 27}\\\\a^3 = \dfrac{2^2\cdot 3^4 + \sqrt{2^4\cdot 3^8-2^2\cdot 3^3\cdot 13^3}}{2\cdot 27}\\\\[/tex]
[tex]a^3 = \dfrac{6\cdot 6\cdot 9+ 6\cdot 5\cdot 7\sqrt{-3}}{6\cdot 9}\\\\a^3 = \dfrac{54+35\sqrt{3} i }{9} = 6 + \dfrac{35}{9}\sqrt{3} \; i\\\\\displaystyle \to a^3-12 = -b^3 = -6 + \frac{\sqrt{35}}{9} \sqrt{3} \; i\\\\b^3 = 6 - \frac{\sqrt{35}}{9} \sqrt{3} \; i[/tex]
[tex]\displaystyle a = \sqrt[3]{6 + \dfrac{35}{9}\sqrt{3} \; i } = \sqrt[6]{6^2+3\cdot \left(\frac{35}{9} \right)^2} \cdot e^{\dfrac{i\cdot \tan^{-1}\left(\displaystyle \frac{35}{54}\sqrt{3} \right)}{3}} \\\\a= \sqrt[6]{\frac{13^3}{3^3} } \cdot e^{\dfrac{i\cdot \tan^{-1}\left(\displaystyle \frac{35}{54}\sqrt{3} \right)}{3}} \\\\[/tex]
[tex]\displaystyle a= \dfrac{\sqrt{39}}{3} \cdot e^{\dfrac{i\cdot \cos^{-1}\left(\displaystyle \frac{54}{\sqrt{6591}} \right)}{3}}\approx \dfrac{\sqrt{39}}{3} \cdot e^{\displaystyle i\cdot \dfrac{4}{45}\pi } \\\\ a\approx 1.92+0.55i\\\\\displaystyle b \approx 1.92-0.55i[/tex]
[tex]\displaystyle x_1 = a+b \approx 4, \\x_2 = a\cdot\frac{-1+\sqrt{3}\cdot i}{2} + b\cdot\frac{-1-\sqrt{3}\cdot i}{2} \approx (-1.44+0.48i)+(-1.44-0.48i)\approx -2.88\\x_2 \approx -3\\\\x_3 = a\cdot\frac{-1-\sqrt{3}\cdot i}{2} + b\cdot\frac{-1+\sqrt{3}\cdot i}{2} \approx -1[/tex]
[tex]\Huge{\boxed{\boxed{\boldsymbol{x = \{-3,-1,-4\}}}}}[/tex]
[answer.2.content]