Akar-akar persamaan kuadrat x² + (a-1)x + 2 = 0 adalah α dan β.
Jika α = 2β dan a > 0 maka nilai a?
Jika α = 2β dan a > 0 maka nilai a?
IcukSugiarto
![\\$Sehingga berikutnya, untuk kasus I (dimana \alpha =2 ; \beta =1):$\\\\\displaystyle \alpha + \beta = \frac{-b}{a}\\\\ 2 + 1 = \frac{-(a-1)}{1} \\\\ 3 = -(a-1)\\ 3 = -a+1\\ a= 1-3\\ a= -2~~~~~~~~~~\Rightarrow [Tidak\ memenuhi\ syarat\ a>0]\\\\\\$Sehingga berikutnya, untuk kasus II (dimana \alpha =-2 ; \beta =-1):$\\\\\displaystyle \alpha + \beta = \frac{-b}{a}\\\\ -2 + (-1) = \frac{-(a-1)}{1} \\\\ -3 = -(a-1)\\ -3 = -a+1\\ a= 1+3\\ a= 4~~~~~~~~~~\Rightarrow [Memenuhi\ syarat\ a>0]](https://tex.z-dn.net/?f=%5C%5C%24Sehingga+berikutnya%2C+untuk+kasus+I+%28dimana++%5Calpha+%3D2+%3B++%5Cbeta+%3D1%29%3A%24%5C%5C%5C%5C%5Cdisplaystyle+%5Calpha+%2B+%5Cbeta+%3D+%5Cfrac%7B-b%7D%7Ba%7D%5C%5C%5C%5C+2+%2B+1+%3D+%5Cfrac%7B-%28a-1%29%7D%7B1%7D+%5C%5C%5C%5C+3+%3D+-%28a-1%29%5C%5C+3+%3D+-a%2B1%5C%5C+a%3D+1-3%5C%5C+a%3D+-2%7E%7E%7E%7E%7E%7E%7E%7E%7E%7E%5CRightarrow+%5BTidak%5C+memenuhi%5C+syarat%5C+a%3E0%5D%5C%5C%5C%5C%5C%5C%24Sehingga+berikutnya%2C+untuk+kasus+II+%28dimana++%5Calpha+%3D-2+%3B++%5Cbeta+%3D-1%29%3A%24%5C%5C%5C%5C%5Cdisplaystyle+%5Calpha+%2B+%5Cbeta+%3D+%5Cfrac%7B-b%7D%7Ba%7D%5C%5C%5C%5C+-2+%2B+%28-1%29+%3D+%5Cfrac%7B-%28a-1%29%7D%7B1%7D+%5C%5C%5C%5C+-3+%3D+-%28a-1%29%5C%5C+-3+%3D+-a%2B1%5C%5C+a%3D+1%2B3%5C%5C+a%3D+4%7E%7E%7E%7E%7E%7E%7E%7E%7E%7E%5CRightarrow+%5BMemenuhi%5C+syarat%5C+a%3E0%5D "\\$Sehingga berikutnya, untuk kasus I (dimana \alpha =2 ; \beta =1):$\\\\\displaystyle \alpha + \beta = \frac{-b}{a}\\\\ 2 + 1 = \frac{-(a-1)}{1} \\\\ 3 = -(a-1)\\ 3 = -a+1\\ a= 1-3\\ a= -2~~~~~~~~~~\Rightarrow [Tidak\ memenuhi\ syarat\ a>0]\\\\\\$Sehingga berikutnya, untuk kasus II (dimana \alpha =-2 ; \beta =-1):$\\\\\displaystyle \alpha + \beta = \frac{-b}{a}\\\\ -2 + (-1) = \frac{-(a-1)}{1} \\\\ -3 = -(a-1)\\ -3 = -a+1\\ a= 1+3\\ a= 4~~~~~~~~~~\Rightarrow [Memenuhi\ syarat\ a>0]")
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