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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

$${solve}: \\ $$$$\left({a}+{b}−\mathrm{2}{c}\right){x}^{\mathrm{2}} +\left(\mathrm{2}{a}−{b}−{c}\right){x}+\left({c}+{a}−\mathrm{2}{b}\right)=\mathrm{0} \\ $$

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))

$$\left({a}+{b}−\mathrm{2}{c}\right){x}^{\mathrm{2}} +\left({a}+{b}−\mathrm{2}{c}\right){x} \\ $$$$\:\:\:+\left({c}+{a}−\mathrm{2}{b}\right){x}+\left({c}+{a}−\mathrm{2}{b}\right)=\mathrm{0} \\ $$$$\left({a}+{b}−\mathrm{2}{c}\right){x}\left({x}+\mathrm{1}\right)+\left({c}+{a}−\mathrm{2}{b}\right)\left({x}+\mathrm{1}\right)=\mathrm{0} \\ $$$$\left({x}+\mathrm{1}\right)\left\{\left({a}+{b}−\mathrm{2}{c}\right){x}+\left({c}+{a}−\mathrm{2}{b}\right)\right\}=\mathrm{0} \\ $$$$\therefore\:{x}=−\mathrm{1} \\ $$$$\:{x}=\frac{\mathrm{2}{b}−{c}−{a}}{{a}+{b}−\mathrm{2}{c}} \\ $$

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