Latihan Matematika Peminatan Kelas XI Akar-Akar Suku Banyak
# 10
Pilgan

Persamaan polinom yang akar-akarnya (23), (2+3) dan 1\left(\sqrt{2}-3\right),\ \left(\sqrt{2}+3\right)\ \text{dan}\ 1 adalah ....

A

x3(1+22)x2+(7+22)x+7=0x^3-\left(1+2\sqrt{2}\right)x^2+\left(-7+2\sqrt{2}\right)x+7=0

B

x3+(1+22)x2(7+22)x7=0x^3+\left(1+2\sqrt{2}\right)x^2-\left(-7+2\sqrt{2}\right)x-7=0

C

x3(23)x2+(23)x+1=0x^3-\left(\sqrt{2}-3\right)x^2+\left(\sqrt{2}-3\right)x+1=0

D

x3(22)x2+(22)x7=0x^3-\left(2\sqrt{2}\right)x^2+\left(2\sqrt{2}\right)x-7=0

E

x3(22+1)x2+(221)x7=0x^3-\left(2\sqrt{2}+1\right)x^2+\left(2\sqrt{2}-1\right)x-7=0

Pembahasan:

Gunakan teorema akar-akar x1+x2+x3=bax_1+x_2+x_3=-\frac{b}{a}

x1+x2+x3\Leftrightarrow x_1+x_2+x_3

(23)+(2+3)+1=1+22\Leftrightarrow\left(\sqrt{2}-3\right)+\left(\sqrt{2}+3\right)+1=1+2\sqrt{2}

Gunakan teorema akar-akar x1×x2+x2×x3+x1×x3=cax_1\times x_2+x_2\times x_3+x_1\times x_3=\frac{c}{a}

 x1×x2+x2×x3+x1×x3\Leftrightarrow\ x_1\times x_2+x_2\times x_3+x_1\times x_3

 (23)× (2+3)+(2+3)× 1+(23)× 1 \Leftrightarrow\ \left(\sqrt{2}-3\right)\times\ \left(\sqrt{2}+3\right)+\left(\sqrt{2}+3\right)\times\ 1+\left(\sqrt{2}-3\right)\times\ 1\

(29)+(2+3)+(23)\Leftrightarrow\left(2-9\right)+\left(\sqrt{2}+3\right)+\left(\sqrt{2}-3\right)

7+22\Leftrightarrow-7+2\sqrt{2}

Gunakan teorema akar-akar x1×x2×x3=dax_1\times x_2\times x_3=-\frac{d}{a}

x1×x2×x3\Leftrightarrow x_1\times x_2\times x_3

(23)×(2+3)×1\Leftrightarrow\left(\sqrt{2}-3\right)\times\left(\sqrt{2}+3\right)\times1

(29)×1=7\Leftrightarrow\left(2-9\right)\times1=-7

Bentuk persamaan umum pangkat tiga adalah ax3+bx2+cx+d=0ax^3+bx^2+cx+d=0

Setelah melakukan perhitungan, maka: a=1, b=(1+22), c=7+22dan d=7a=1,\ b=-\left(1+2\sqrt{2}\right),\ c=-7+2\sqrt{2}\text{dan}\ d=7

Jadi, persamaannya adalah x3(1+22)x2+(7+22)x+7=0x^3-\left(1+2\sqrt{2}\right)x^2+\left(-7+2\sqrt{2}\right)x+7=0

1 2 3 4 5 6 7 8 9 10