Question

1. Selesaikan persamaan [tex]x^{2} +\frac{x^{2} }{(x+1)^{2} } = 3[/tex]
2. Tentukan 2 akar real [tex]2000x^{6} +100x^{5} +10x^{3} +x-2=0[/tex]

Answer 1

1. [tex] x_{1} = \frac{1-\sqrt{5}}{2} \: \text{ dan } \: x_{2} = \frac{1+\sqrt{5}}{2} [/tex] .

2. [tex] x_{1} = \frac{-1-\sqrt{161}}{40} \: \text{ dan } \: x_{2} = \frac{-1+\sqrt{161}}{40} [/tex] .

Pembahasan

1.

[tex] \begin{aligned} x^2+\frac{x^2}{(x+1)^2} & \: = 3 \\ \\ \frac{(x(x+1))^2+x^2}{(x+1)^2} \: & = 3 \\ \\ (x(x+1))^2+x^2 \: & = 3(x+1)^2 \\ \\ x^4+2x^3+2x^2 \: & = 3x^2+6x+3 \\ \\ \Leftrightarrow \: \: x^4+2x^3-x^2-6x-3 \: & = 0 \\ \\ \left(x^2+3x+3\right)\left(x^2-x-1\right) \: & = 0 \\ \\ \end{aligned} [/tex]

[tex] \begin{aligned} x^2+3x+3 & \: = 0 \\ \\ D \: & = 3^2-4\cdot 1\cdot 3 \\ \\ D \: & = 9-12 \\ \\ D \: & = -3 \\ \\ \end{aligned} [/tex]

Memiliki akar-akar imajiner ( tidak real ) .

[tex] \begin{aligned} x^2-x-1 & \: = 0 \\ \\ D \: & = (-1)^2-4\cdot 1\cdot (-1) \\ \\ D \: & = 1+4 \\ \\ D \: & = 5 \\ \\ \end{aligned} [/tex]

Memiliki akar-akar real dan berbeda.

[tex] \begin{aligned} x^2-x-1 & \: = 0 \\ \\ \left(x^2-x\right) \: & = 1 \\ \\ \left(x-\frac{1}{2}\right)^2 - \frac{1}{4} \: & = 1 \\ \\ \left(x-\frac{1}{2}\right)^2 \: & = 1 + \frac{1}{4} \\ \\ \left(x-\frac{1}{2}\right)^2 \: & = \frac{5}{4} \\ \\ x-\frac{1}{2} \: & = \pm \sqrt{\frac{5}{4}} \\ \\ x \: & = \frac{1}{2} \pm \sqrt{\frac{5}{4}} \\ \\ x \: & = \frac{1}{2} \pm \frac{\sqrt{5}}{2} \\ \\ x \: & = \frac{1 \pm \sqrt{5}}{2} \\ \\ \end{aligned} [/tex]

[tex] x_{1} = \frac{1-\sqrt{5}}{2} \: \text{ dan } \: x_{2} = \frac{1+\sqrt{5}}{2} [/tex] .

2.

[tex] \begin{aligned} 2000x^6+100x^5+10x^3+x-2 & \: = 0 \\ \\ 2000x^6+100x^5-200x^4+200x^4+10x^3-20x^2+20x^2+x-2 \: & = 0 \\ \\ (2000x^6+100x^5-200x^4)+(200x^4+10x^3-20x^2)+(20x^2+x-2) \: & = 0 \\ \\ 100x^4(20x^2+x-2) + 10x^2(20x^2+x-2) + (20x^2+x-2) \: & = 0 \\ \\ \left(100x^4+10x^2+1\right)\left(20x^2+x-2\right) \: & = 0 \\ \\ \end{aligned} [/tex]

[tex] \begin{aligned} 100x^4+10x^2+1 & \: = 0 \\ \\ D \: & = 10^2-4\cdot 100 \cdot 1 \\ \\ D \: & = 100-400 \\ \\ D \: & = -300 \\ \\ \end{aligned} [/tex]

Memiliki akar-akar imajiner ( tidak real ) .

[tex] \begin{aligned} 20x^2+x-2 & \: = 0 \\ \\ D \: & = 1^2-4\cdot 20 \cdot (-2) \\ \\ D \: & = 1+160 \\ \\ D \: & = 161 \\ \\ \end{aligned} [/tex]

Memiliki akar-akar real dan berbeda.

[tex] \begin{aligned} 20x^2+x-2 & \: = 0 \\ \\ x \: & = \frac{-1\pm \sqrt{1^2-4\cdot 20 \cdot (-2)}}{2\cdot 20} \\ \\ x \: & = \frac{-1\pm \sqrt{161}}{40} \\ \\ \end{aligned} [/tex]

[tex] x_{1} = \frac{-1-\sqrt{161}}{40} \: \text{ dan } \: x_{2} = \frac{-1+\sqrt{161}}{40} [/tex] .

Kesimpulan :

1. [tex] x_{1} = \frac{1-\sqrt{5}}{2} \: \text{ dan } \: x_{2} = \frac{1+\sqrt{5}}{2} [/tex] .

2. [tex] x_{1} = \frac{-1-\sqrt{161}}{40} \: \text{ dan } \: x_{2} = \frac{-1+\sqrt{161}}{40} [/tex] .