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Question Number 172511 by Mikenice last updated on 28/Jun/22
solve: (a+b−2c)x^2 +(2a−b−c)x+(c+a−2b)=0
solve:(a+b−2c)x2+(2a−b−c)x+(c+a−2b)=0
Answered by som(math1967) last updated on 28/Jun/22
(a+b−2c)x^2 +(a+b−2c)x +(c+a−2b)x+(c+a−2b)=0 (a+b−2c)x(x+1)+(c+a−2b)(x+1)=0 (x+1){(a+b−2c)x+(c+a−2b)}=0 ∴ x=−1 x=((2b−c−a)/(a+b−2c))
(a+b−2c)x2+(a+b−2c)x+(c+a−2b)x+(c+a−2b)=0(a+b−2c)x(x+1)+(c+a−2b)(x+1)=0(x+1){(a+b−2c)x+(c+a−2b)}=0∴x=−1x=2b−c−aa+b−2c

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