A. Akar-Akar Persamaan Polinomial Berderajat 3
Diketahui \(ax^3 + bx^2 + cx + d = 0\) memiliki akar-akar \(x_1, x_2, \text{ dan } x_3\)
- \(x_1 + x_2 + x_3 = -\dfrac{b}{a}\)
- \(x_1\cdot x_2 + x_1 \cdot x_3 + x_2 \cdot x_3 = \dfrac{c}{a}\)
- \(x_1 \cdot x_2 \cdot x_3 = -\dfrac{d}{a}\)
Contoh:
Diketahui persamaan polinomial \(x^3 + 2x^2 \:-\: 3x + 5 = 0\) memiliki akar-akar \(x_1, x_2, \text{ dan } x_3\)
Tentukan:
(a) \(x_1 + x_2 + x_3 \)
(b) \(x_1\cdot x_2 + x_1 \cdot x_3 + x_2 \cdot x_3\)
(c) \(x_1 \cdot x_2 \cdot x_3\)
Penyelesaian:
\(x^3 + 2x^2 \:-\: 3x + 5 = 0\)
\(\color{red} a = 1, b = 2, c = -3, d = 5\)
(a) \(x_1 + x_2 + x_3 = -\dfrac{b}{a}=-\dfrac{2}{1} = -2\)
(b) \(x_1\cdot x_2 + x_1 \cdot x_3 + x_2 \cdot x_3 = \dfrac{c}{a}=\dfrac{-3}{1} = -3\)
(c) \(x_1 \cdot x_2 \cdot x_3 = -\dfrac{d}{a}=-\dfrac{5}{1} = -5\)
B. Akar-Akar Persamaan Polinomial Berderajat 4
Diketahui \(ax^4 + bx^3 + cx^2 + dx + c = 0\) memiliki akar-akar \(x_1, x_2, x_3, \text{ dan } x_4\)
- \(x_1 + x_2 + x_3 + x_4 = -\dfrac{b}{a}\)
- \(x_1 \cdot x_2 + x_1 \cdot x_3 + x_1 \cdot x_4 + x_2 \cdot x_3 + x_2 \cdot x_4 + x_3 \cdot x_4 = \dfrac{c}{a}\)
- \(x_1\cdot x_2 \cdot x_3 + x_1 \cdot x_2 \cdot x_4 + x_1 \cdot x_3 \cdot x_4 + x_2 \cdot x_3 \cdot x_4 = – \dfrac{d}{a}\)
- \(x_1 \cdot x_2 \cdot x_3 \cdot x_4 = \dfrac{e}{a}\)
Contoh:
Diketahui persamaan polinomial \(x^4 \:-\:5x^3 \:-\: 3x^2 + 5x\:-\:10 = 0\) memiliki akar-akar \(x_1, x_2, x_3\text{ dan } x_4\)
Tentukan:
(a) \(x_1 + x_2 + x_3 + x_4 \)
(b) \(x_1 \cdot x_2 + x_1 \cdot x_3 + x_1 \cdot x_4 + x_2 \cdot x_3 + x_2 \cdot x_4 + x_3 \cdot x_4\)
(c) \(x_1\cdot x_2 \cdot x_3 + x_1 \cdot x_2 \cdot x_4 + x_1 \cdot x_3 \cdot x_4 + x_2 \cdot x_3 \cdot x_4\)
(d) \(x_1 \cdot x_2 \cdot x_3 \cdot x_4\)
Penyelesaian:
\(x^4 \:-\:5x^3 \:-\: 3x^2 + 5x\:-\:10 = 0\)
\(\color{red} a = 1, b = -5, c = -3, d = 5, e = -10\)
(a) \(x_1 + x_2 + x_3 + x_4 = -\dfrac{b}{a} = -\dfrac{-5}{1} = 5\)
(b) \(x_1 \cdot x_2 + x_1 \cdot x_3 + x_1 \cdot x_4 + x_2 \cdot x_3 + x_2 \cdot x_4 + x_3 \cdot x_4 = \dfrac{c}{a}= \dfrac{-3}{1}=-3\)
(c) \(x_1\cdot x_2 \cdot x_3 + x_1 \cdot x_2 \cdot x_4 + x_1 \cdot x_3 \cdot x_4 + x_2 \cdot x_3 \cdot x_4 = – \dfrac{d}{a}=-\dfrac{5}{1} = -5\)
(d) \(x_1 \cdot x_2 \cdot x_3 \cdot x_4 = \dfrac{e}{a}=\dfrac{-10}{1}=-10\)
LATIHAN SOAL
Soal 1
Jika \(\alpha, \beta, \gamma\) merupakan akar-akar polinomial \(3x^3 \:-\:9x^2 + 6x + 2= 0\) maka nilai \((\alpha^2 + \beta^2 + \gamma^2)\left(\dfrac{1}{\alpha} + \dfrac{1}{\beta} + \dfrac{1}{\gamma}\right) = \dotso\)
(A) \(-18\)
(B) \(-15\)
(C) \(15\)
(D) \(18\)
(E) \(20\)
Answer: B
\(\alpha^2 + \beta^2 + \gamma^2 = (\alpha + \beta + \gamma)^2\:-\:2(\alpha\cdot \beta + \alpha \cdot \gamma + \beta \cdot \gamma)\)
\(\alpha^2 + \beta^2 + \gamma^2 = \left(-\dfrac{b}{a}\right)^2\:-\:2(\dfrac{c}{a})\)
\(\alpha^2 + \beta^2 + \gamma^2 = \left(-\dfrac{-9}{3}\right)^2\:-\:2(\dfrac{6}{3})\)
\(\alpha^2 + \beta^2 + \gamma^2 = 9\:-\:4 = 5\)
\(\dfrac{1}{\alpha} + \dfrac{1}{\beta} + \dfrac{1}{\gamma}= \dfrac{\beta \cdot \gamma + \alpha \cdot \gamma + \alpha\cdot \beta}{\alpha \cdot \beta \cdot \gamma}\)
\(\dfrac{1}{\alpha} + \dfrac{1}{\beta} + \dfrac{1}{\gamma}= \dfrac{\dfrac{c}{a}}{-\dfrac{d}{a}}\)
\(\dfrac{1}{\alpha} + \dfrac{1}{\beta} + \dfrac{1}{\gamma}= \dfrac{\dfrac{6}{3}}{-\dfrac{2}{3}}\)
\(\dfrac{1}{\alpha} + \dfrac{1}{\beta} + \dfrac{1}{\gamma}= \dfrac{2}{-\dfrac{2}{3}} = -3\)
\((\alpha^2 + \beta^2 + \gamma^2)\left(\dfrac{1}{\alpha} + \dfrac{1}{\beta} + \dfrac{1}{\gamma}\right) = 5 (-3) = -15\)
Soal 2
Jika \(x_1, x_2, \text{ dan } x_3\) adalah akar-akar persamaan \(2x^3\:-\:8x^2 + 4x \:-\:1 = 0\) maka nilai \(x_1^3\cdot x_2 \cdot x_3 + x_1 \cdot x_2^3 \cdot x_3 + x_1 \cdot x_2 \cdot x_3^3\) adalah…
(A) 3
(B) 6
(C) 12
(D) 16
(E) 18
Answer: B
\(x_1^3\cdot x_2 \cdot x_3 + x_1 \cdot x_2^3 \cdot x_3 + x_1 \cdot x_2 \cdot x_3^3 = x_1 \cdot x_2 \cdot x_3 (x_1^2 + x_2^2 + x_3^2)\)
\(\color{blue}x_1 \cdot x_2 \cdot x_3 = -\dfrac{d}{a} = -\dfrac{-1}{2} = \dfrac{1}{2}\)
\(\color{blue}x_1^2 + x_2^2 + x_3^2 = (x_1 + x_2 + x_3)^2\:-\:2(x_1 \cdot x_2 + x_1 \cdot x_3 + x_2 \cdot x_3)\)
\(\color{blue} x_1^2 + x_2^2 + x_3^2 = \left(-\dfrac{b}{a}\right)^2\:-\:2\left(\dfrac{c}{a}\right)\)
\(\color{blue}x_1^2 + x_2^2 + x_3^2 = \left(-\dfrac{-8}{2}\right)^2\:-\:2\left(\dfrac{4}{2}\right)\)
\(\color{blue}x_1^2 + x_2^2 + x_3^2 = 16\:-\:4 = 12\)
\(x_1^3\cdot x_2 \cdot x_3 + x_1 \cdot x_2^3 \cdot x_3 + x_1 \cdot x_2 \cdot x_3^3 = \dfrac{1}{2} (12) = 6\)
Soal 3
Jika \(x_1, x_2, \text{ dan } x_3\) merupakan akar-akar polinomial \(x^3 \:-\:5x^2 + x + 2 = 0\) maka nilai \(x_1^3 + x_2^3 + x_3^3 = \dotso\)
(A) 104
(B) 114
(C) 126
(D) 130
(E) 133
Answer: A
\(\color{blue} x_1^3 + x_2^3 + x_3^3 \:-\:3x_1x_2x_3 = (x_1 + x_2 + x_3)(x_1^2 + x_2^2 + x_3^2\:-\:(x_1x_2 + x_1x_3 + x_2x_3))\)
\(\color{purple} x_1^2 + x_2^2 + x_3^2 = (x_1 + x_2 + x_3)^2\:-\:2(x_1x_2 + x_1x_3 + x_2x_3)\)
\(\color{purple} x_1^2 + x_2^2 + x_3^2 = \left(-\dfrac{b}{a}\right)^2\:-\:2\left(\dfrac{c}{a}\right)\)
\(x^3 \:-\:5x^2 + x + 2 = 0\)
\(a = 1, b = -5, c = 1, d = 2\)
\(\color{purple} x_1^2 + x_2^2 + x_3^2 = \left(-\dfrac{(-5)}{1}\right)^2\:-\:2\left(\dfrac{1}{1}\right) = 23\)
\(x_1^3 + x_2^3 + x_3^3 \:-\:3x_1x_2x_3 = (x_1 + x_2 + x_3)(x_1^2 + x_2^2 + x_3^2\:-\:(x_1x_2 + x_1x_3 + x_2x_3))\)
\(x_1^3 + x_2^3 + x_3^3 \:-\:3(-2) = (5)(23\:-\:(1))\)
\(x_1^3 + x_2^3 + x_3^3 + 6 = 110\)
\(x_1^3 + x_2^3 + x_3^3 = 104\)
Soal 4
Diketahui \(x_1 + x_2 + x_3 = \dfrac{3}{2}\), \(x_1^2 + x_2^2 + x_3^2 = \dfrac{1}{4}\) dan \(x_1^3 + x_2^3 + x_3^3 = -\dfrac{33}{8}\). Persamaan suku banyak tersebut adalah…
(A) \(2x^3\:-\:3x^2 + 2x + 2\)
(B) \(2x^3 + 3x^2 + 2x + 2\)
(C) \(2x^3\:-\:3x^2 \:-\: 2x + 2\)
(D) \(3x^3\:-\:3x^2 + 2x + 3\)
(E) \(4x^3\:-\:3x^2 + 3x + 2\)
Answer: A
Persamaan polinomial yang memiliki 3 buah akar adalah persamaan polinomial tingkat 3, yaitu: \(ax^3 + bx^2 + cx + d = 0\)
\(x_1 + x_2 + x_3 = \dfrac{3}{2} = -\dfrac{b}{a}\)
\(a = 2 \text{ dan } b = -3\)
\(x_1^2 + x_2^2 + x_3^2 = (x_1 + x_2 + x_3)^2\:-\:2(x_1x_2 + x_1x_3 + x_2x_3)\)
\(\dfrac{1}{4} = \left(\dfrac{3}{2} \right)^2 \:-\:2\left(\dfrac{c}{a} \right)\)
\(\dfrac{1}{4} = \dfrac{9}{4} \:-\:2\left(\dfrac{c}{a} \right)\)
\(\dfrac{8}{4} = \dfrac{2c}{2}\)
\(c = 2\)
\(x_1^3 + x_2^3 + x_3^3 \:-\:3x_1x_2x_3 = (x_1 + x_2 + x_3)(x_1^2 + x_2^2 + x_3^2\:-\:(x_1x_2 + x_1x_3 + x_2x_3))\)
\(-\dfrac{33}{8}\:-\:3\left(-\dfrac{d}{a}\right) = \dfrac{3}{2}\left(\dfrac{1}{4}\:-\:\dfrac{c}{a}\right)\)
\(-\dfrac{33}{8} + \dfrac{3d}{2} = \dfrac{3}{2}\left(\dfrac{1}{4}\:-\:\dfrac{2}{2}\right)\)
\(-\dfrac{33}{8} + \dfrac{3d}{2} = -\dfrac{9}{8}\)
\(\dfrac{3d}{2} = 3\)
\(d = 2\)
Jadi diperoleh \(a = 2, b = -3, c = 2, d = 2\)
Suku banyak tersebut adalah \(2x^3 \:-\:3x^2 + 2x + 2\)
Soal 5
Diketahui \(x^3 \:-\:2x^2 + 2 = 0\) dengan akar-akar \(x_1, x_2, \text{ dan } x_3\). Nilai \(x_1^6x_2^2 x_3^2 + x_1^2 x_2^6 x_3^2 + x_1^2x_2^2x_3^6 = \dotso\)
(A) 0
(B) 1
(C) 2
(D) 3
(E) 4
Answer: A
\(x^3 \:-\:2x^2 + 2 = 0\)
\(x_1 + x_2 + x_ 3 = -\dfrac{b}{a} = 2\)
\(x_1x_2 + x_1x_3 + x_2x_3 = \dfrac{c}{a} = 0\)
\(x_1x_2x_3 = -\dfrac{d}{a} = -2\)
\(x_1^6x_2^2 x_3^2 + x_1^2 x_2^6 x_3^2 + x_1^2x_2^2x_3^6 = x_1^2x_2^2x_3^2(x_1^4 + x_2^4 + x_3^4)\)
\(\color{blue} x_1^4 + x_2^4 + x_3^4 = (x_1^2 + x_2^2 + x_3^2)^2\:-\:2[(x_1x_2 + x_1x_3 + x_2x_3)^2 \:-\:2x_1x_2x_3(x_1 + x_2 + x_3)]\)
\(x_1^4 + x_2^4 + x_3^4 = 4^2\:-\:2(0^2\:-\:2(-2)(2))\)
\(x_1^4 + x_2^4 + x_3^4 = 0\)
\(x_1^6x_2^2 x_3^2 + x_1^2 x_2^6 x_3^2 + x_1^2x_2^2x_3^6 = (x_1x_2x_3)^2\cdot (x_1^4 + x_2^4 + x_3^4)\)
\(x_1^6x_2^2 x_3^2 + x_1^2 x_2^6 x_3^2 + x_1^2x_2^2x_3^6 = (-2)^2\cdot (0) = 0\)
