Respostas

2014-06-15T18:11:53-03:00
(K-2)x^{2}-3kx+ 1 = 0, com K\neq2
x'+x"= \frac{-b}{a}  \\ x'.x"= \frac{c}{a}
x'+x"= \frac{-b}{a}  \\ x'+x"= \frac{-(-3k)}{k-2}
x'.x"=\frac{c}{a}  \\x'.x"= \frac{1}{k-2}
x'.x"=x'+x"
\frac{(3k)}{k-2} = \frac{1}{k-2}
multiplicando cruzado
3k({k-2}) ={k-2} \\ 3k^{2}-6k={k-2} \\3k^{2}-7k+2=0
Agora você resolve baskara para achar o K
k'=2 ou k"= \frac{1}{3}


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