Jawaban:
kan 3×+6×-1=10 berarti persamaannya ituu ada di 3×2+6×-6=0
Jawab:
3x² - 18x + 11 = 0
Penjelasan dengan langkah-langkah:
3x² + 6x - 1 = 0 ← bentuk ax² + bx + c = 0
[tex]\displaystyle \alpha+\beta=-\frac{b}{a}=-\frac{6}{3}=-2\\\alpha\beta=\frac{c}{a}=\frac{-1}{3}[/tex]
Misal akar-akar persamaan kuadrat baru nya x₁ atau x₂, maka:
[tex]\displaystyle x^2-(x_1+x_2)x+x_1x_2=0\\x^2-(1-2\alpha+1-2\beta)x+(1-2\alpha)(1-2\beta)=0\\x^2-2[1-(\alpha+\beta)]x+[1-2(\alpha+\beta)+4\alpha\beta]=0\\x^2-2[1-(-2)]x+\left [ 1-2(-2)+4\left ( -\frac{1}{3} \right ) \right ]=0\\x^2-6x+\frac{11}{3}=0\\3x^2-18x+11=0[/tex]
" Life is not a problem to be solved but a reality to be experienced! "
© Copyright 2013 - 2024 KUDO.TIPS - All rights reserved.
Jawaban:
kan 3×+6×-1=10 berarti persamaannya ituu ada di 3×2+6×-6=0
Jawab:
3x² - 18x + 11 = 0
Penjelasan dengan langkah-langkah:
3x² + 6x - 1 = 0 ← bentuk ax² + bx + c = 0
[tex]\displaystyle \alpha+\beta=-\frac{b}{a}=-\frac{6}{3}=-2\\\alpha\beta=\frac{c}{a}=\frac{-1}{3}[/tex]
Misal akar-akar persamaan kuadrat baru nya x₁ atau x₂, maka:
[tex]\displaystyle x^2-(x_1+x_2)x+x_1x_2=0\\x^2-(1-2\alpha+1-2\beta)x+(1-2\alpha)(1-2\beta)=0\\x^2-2[1-(\alpha+\beta)]x+[1-2(\alpha+\beta)+4\alpha\beta]=0\\x^2-2[1-(-2)]x+\left [ 1-2(-2)+4\left ( -\frac{1}{3} \right ) \right ]=0\\x^2-6x+\frac{11}{3}=0\\3x^2-18x+11=0[/tex]