Question Number 172511 by Mikenice last updated on 28/Jun/22
$${solve}: \\ $$$$\left({a}+{b}−\mathrm{2}{c}\right){x}^{\mathrm{2}} +\left(\mathrm{2}{a}−{b}−{c}\right){x}+\left({c}+{a}−\mathrm{2}{b}\right)=\mathrm{0} \\ $$
Answered by som(math1967) last updated on 28/Jun/22
$$\left({a}+{b}−\mathrm{2}{c}\right){x}^{\mathrm{2}} +\left({a}+{b}−\mathrm{2}{c}\right){x} \\ $$$$\:\:\:+\left({c}+{a}−\mathrm{2}{b}\right){x}+\left({c}+{a}−\mathrm{2}{b}\right)=\mathrm{0} \\ $$$$\left({a}+{b}−\mathrm{2}{c}\right){x}\left({x}+\mathrm{1}\right)+\left({c}+{a}−\mathrm{2}{b}\right)\left({x}+\mathrm{1}\right)=\mathrm{0} \\ $$$$\left({x}+\mathrm{1}\right)\left\{\left({a}+{b}−\mathrm{2}{c}\right){x}+\left({c}+{a}−\mathrm{2}{b}\right)\right\}=\mathrm{0} \\ $$$$\therefore\:{x}=−\mathrm{1} \\ $$$$\:{x}=\frac{\mathrm{2}{b}−{c}−{a}}{{a}+{b}−\mathrm{2}{c}} \\ $$