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OSP SMA 2007 Bagian Kedua No 3

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Tentukan semua bilangan real $x$ yang memenuhi $x^4 - 4x^3 + 5x^2 - 4x + 1 = 0$.

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$x^4-4x^3+5x^2-4x+1=0$
$x^4-4x^3+6x^2-4x+1-x^2=0$

$(x-1)^4-x^2=0$

$(x^2-x+1)(x^2-3x+1)=0$

$x^2-x+1=0$ atau $x^2-3x+1=0$

Perhatikan bahwa $x^2-x+1=0$ tidak memiliki akar real karena $D=(-1)^2-4(1)(1)=-3<0$

Sehingga yang mempunyai akar real hanya $x^2-3x+1=0$

yaitu,

$$x_1,2 = \frac{-b \ps \sqrt{b^2-4ac}}{2a}$$

$$=\frac{3 \ps \sqrt{(-3)^2-4(1)(1)}}{2}=\frac{3 \ps\sqrt5}{2}$$

$x_1=\frac{3+\sqrt5}{2}$ dan $x_2=\frac{3-\sqrt5}{2}$

Edited by Lucas Lawrence

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