QUIZ
Tentukan nilai x, jika x adalah bilangan real
[tex] \frac{ {x}^{11} + x}{ {x}^{7} + {x}^{5} } = \frac{205}{16} [/tex]
Tentukan nilai x, jika x adalah bilangan real
[tex] \frac{ {x}^{11} + x}{ {x}^{7} + {x}^{5} } = \frac{205}{16} [/tex]
(x¹¹ + x) / (x⁷ + x⁵) = 205/16
x(x¹⁰ + 1) / x⁵(x² + 1) = 205/16
(x¹⁰ + 1) / x⁵. x / (x² + 1) = 205/16
x⁵ + 1/x⁵ = 205/16 . (x² + 1) / x
x⁵ + 1/x⁵ = 205/16 . x + 1/x
misalkan..
x + 1/x = a
mencari nilai x⁵ + 1/x⁵ :
(x + 1/x)⁵ = x⁵ + 1/x⁵ + 5x³ + 10x + 10/x + 5/x³
a⁵ = x⁵ + 1/x⁵ + 10x + 10/x + 5x³ + 5/x³
a⁵ = x⁵ + 1/x⁵ + 10(x + 1/x) + 5(x³ + 1/x³)
a⁵ = x⁵ + 1/x⁵ + 10a + 5(x³ + 1/x³)
mencari nilai x³ + 1/x³ :
(x + 1/x)³ = x³ + 1/x³ + 3x(1/x) (x + 1/x)
a³ = x³ + 1/x³ + 3(x + 1/x)
a³ = x³ + 1/x³ + 3a
x³ + 1/x³ = a³ - 3a
subsitusikan nilai x³ + 1/x³...
a⁵ = x⁵ + 1/x⁵ + 10a + 5(x³ + 1/x³)
a⁵ = x⁵ + 1/x⁵ + 10a + 5(a³ - 3a)
a⁵ = x⁵ + 1/x⁵ + 10a + 5a³ - 15a
a⁵ = x⁵ + 1/x⁵ - 5a + 5a³
x⁵ + 1/x⁵ = a⁵ - 5a³ + 5a
subsitusikan nilai x⁵ + 1/x⁵ ke persamaan yang awal..
x⁵ + 1/x⁵ = 205/16 . x + 1/x
a⁵ - 5a³ + 5a = 205/16 . a
16(a⁵ - 5a³ + 5a) = 205a
16a⁵ - 80a³ + 80a = 205a
16a⁵ - 80a³ - 125a = 0
a(16a⁴ - 80a² - 125) = 0
a = 0
16a⁴ - 80a² - 125 = 0
(4a²)² - 20(4a²) - 125 = 0
misalkan..
4a² = u
u² - 20u - 125 = 0
(u + 5) (u - 25) = 0
u1 = -5
u2 = 25
karena u1 bernilai negatif, maka solusinya hanya u2.
4a² = u
4a² = 25
a² = 25/4
a = ± 5/2
mencari nilai x, jika a = 5/2
x + 1/x = a
x + 1/x = 5/2
2(x + 1/x) = 5
2x + 2/x = 5
2x² + 2 = 5x
2x² - 5x + 2 = 0
rumus kuadratik :
x = -b ± √b² - 4ac / 2a
x = 5 ± √-5² - 4(2)(2) / 2(2)
x = 5 ± √25 - 16 / 4
x = 5 ± √9 / 4
x = 5 ± 3 / 4
x1 = 5 + 3 / 4
x1 = 8/4
x1 = 2
x2 = 5 - 3 / 4
x2 = 2/4
x2 = 1/2
mencari nilai x, jika a = -5/2
x + 1/x = a
x + 1/x = -5/2
2(x + 1/x) = -5
2x + 2/x = -5
2x² + 2 = -5x
2x² + 5x + 2 = 0
rumus kuadratik :
x = -b ± √b² - 4ac / 2a
x = -5 ± √5² - 4(2)(2) / 2(2)
x = -5 ± √25 - 16 / 4
x = -5 ± √9 / 4
x = -5 ± 3 / 4
x3 = -5 + 3 / 4
x3 = -2/4
x3 = -1/2
x4 = -5 - 3 / 4
x4 = -8/4
x4 = -2
maka, nilai x adalah 2, 1/2, -1/2, dan -2