considerando a equação 12x^2-(m+2)x-1=0 em que a soma das raizes é igual a 5/6. Nessas condições, determine o valor de m:
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12x² -(m+2)x -1=0
a=12
b=-(m+2)
c=-1
![Soma= \frac{-b}{a} \\ \\ \frac{5}{6} = \frac{-[-(m+2)]}{12} \\ \\ \frac{5}{6} = \frac{(m+2)}{12} \\ \\ 6.(m+2)=5.12 \\ \\ 6m+12=60 \\ \\ 6m=60-12 \\ \\ 6m=48 \\ \\ m=48\div6 \\ \\ m=8](https://tex.z-dn.net/?f=Soma%3D++%5Cfrac%7B-b%7D%7Ba%7D++%5C%5C++%5C%5C++%5Cfrac%7B5%7D%7B6%7D+%3D+%5Cfrac%7B-%5B-%28m%2B2%29%5D%7D%7B12%7D++%5C%5C++%5C%5C++%5Cfrac%7B5%7D%7B6%7D+%3D+%5Cfrac%7B%28m%2B2%29%7D%7B12%7D++%5C%5C++%5C%5C+6.%28m%2B2%29%3D5.12+%5C%5C++%5C%5C+6m%2B12%3D60+%5C%5C++%5C%5C+6m%3D60-12+%5C%5C++%5C%5C+6m%3D48+%5C%5C++%5C%5C+m%3D48%5Cdiv6+%5C%5C++%5C%5C+m%3D8 "Soma= \frac{-b}{a} \\ \\ \frac{5}{6} = \frac{-[-(m+2)]}{12} \\ \\ \frac{5}{6} = \frac{(m+2)}{12} \\ \\ 6.(m+2)=5.12 \\ \\ 6m+12=60 \\ \\ 6m=60-12 \\ \\ 6m=48 \\ \\ m=48\div6 \\ \\ m=8")
a=12
b=-(m+2)
c=-1
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