solve: (a+b−2c)x^2 +(2a−b−c)x+(c+a−2b)=0


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Question Number 172511 by Mikenice last updated on 28/Jun/22

$s o l v e :$ $(a + b - 2 c) x^{2} + (2 a - b - c) x + (c + a - 2 b) = 0$
	Answered by som(math1967) last updated on 28/Jun/22
	$(a+b−2c)x^2 +(a+b−2c)x +(c+a−2b)x+(c+a−2b)=0 (a+b−2c)x(x+1)+(c+a−2b)(x+1)=0 (x+1){(a+b−2c)x+(c+a−2b)}=0 ∴ x=−1 x=((2b−c−a)/(a+b−2c))$
	$(a + b - 2 c) x^{2} + (a + b - 2 c) x$ $+ (c + a - 2 b) x + (c + a - 2 b) = 0$ $(a + b - 2 c) x (x + 1) + (c + a - 2 b) (x + 1) = 0$ $(x + 1) {(a + b - 2 c) x + (c + a - 2 b)} = 0$ $∴ x = - 1$ $x = \frac{2 b - c - a}{a + b - 2 c}$
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