Tentukan akar-akar dari persamaan dibawah
[tex]\displaystyle \frac{1}{x^2-10x-29}+\frac{1}{x^2-10x-45}+\frac{2}{x^2-10x-69}=0[/tex]
[tex]\displaystyle \frac{1}{x^2-10x-29}+\frac{1}{x^2-10x-45}+\frac{2}{x^2-10x-69}=0[/tex]
Verified answer
PEMBAHASAN
Misal :
x² - 10x - 39 = n
x² - 10x - 29 = n + 10
x² - 10x - 45 = n - 6
x² - 10x - 69 = n - 30
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[tex]\displaystyle \frac{1}{x^2-10x-29}+\frac{1}{x^2-10x-45}+\frac{2}{x^2-10x-69}=0 \\ \\ \frac{1}{n + 10} + \frac{1}{n - 6} - \frac{2}{n - 30} = 0[/tex]
kedua ruas kalikan (n + 10)(n - 6)(n + 30)
(n - 6)(n - 30) + (n + 10)(n - 30) - 2(n + 10)(n - 6) = 0
n² - 36n + 180 + n² - 20n - 300 - 2n² - 8n + 120 = 0
-64n = 0
n = 0
x² - 10x - 39 = 0
(x + 3)(x - 13) = 0
x = -3 atau x = 13
Akar-akar persamaan adalah (-3) dan 13.