Question Number 220853 by fantastic last updated on 20/May/25
$${The}\:{two}\:{solutions}\:{of}\:{the}\:{equation}\:{are}\:{the}\:{same} \\ $$$${a}\left({b}−{c}\right){x}^{\mathrm{2}\:} +{b}\left({c}−{a}\right){x}+{c}\left({a}−{b}\right)=\mathrm{0} \\ $$$${Prove}\:{that}\:\frac{\mathrm{1}}{{a}}+\frac{\mathrm{1}}{{c}}=\frac{\mathrm{2}}{{b}} \\ $$
Answered by fantastic last updated on 20/May/25
$${In}\:{any}\:{quadratic}\:{equation}\:\alpha{x}^{\mathrm{2}} +\beta{x}+{c}=\mathrm{0} \\ $$$${x}=\frac{−\beta\pm\sqrt{\beta^{\mathrm{2}} −\mathrm{4}\alpha{c}}}{\mathrm{2}\alpha}\:.{Now}\:{the}\:{two}\:{solutions}\:{will}\:{be}\:{the}\:{same}\:{if}\:\sqrt{\beta^{\mathrm{2}} −\mathrm{4}\alpha{c}}=\mathrm{0} \\ $$$${So}\: \\ $$$$\left\{{b}\left({c}−{a}\right)\right\}^{\mathrm{2}} \:−\mathrm{4}.{a}\left({b}−{c}\right).{c}\left({a}−{b}\right)=\mathrm{0}\: \\ $$$${or}\:{b}^{\mathrm{2}} \left({c}^{\mathrm{2}} −\mathrm{2}{ac}+{a}^{\mathrm{2}} \right)−\mathrm{4}{ac}\left({ab}−{ac}−{b}^{\mathrm{2}} +{bc}\right)=\mathrm{0} \\ $$$${or}\:{b}^{\mathrm{2}} {c}^{\mathrm{2}} −\mathrm{2}{acb}^{\mathrm{2}} +{a}^{\mathrm{2}} {b}^{\mathrm{2}} −\mathrm{4}{a}^{\mathrm{2}} {bc}+\mathrm{4}{a}^{\mathrm{2}} {c}^{\mathrm{2}} +\mathrm{4}{acb}^{\mathrm{2}} −\mathrm{4}{abc}^{\mathrm{2}} =\mathrm{0} \\ $$$${or}\:{b}^{\mathrm{2}} {c}^{\mathrm{2}} +{a}^{\mathrm{2}} {b}^{\mathrm{2}} +\mathrm{4}{a}^{\mathrm{2}} {c}^{\mathrm{2}} +\mathrm{2}{acb}^{\mathrm{2}} −\mathrm{4}{a}^{\mathrm{2}} {bc}−\mathrm{4}{abc}^{\mathrm{2}} =\mathrm{0} \\ $$$${or}\:\left({bc}+{ab}\:−\mathrm{2}{ac}\right)^{\mathrm{2}} =\mathrm{0} \\ $$$${or}\:{bc}+{ab}−\mathrm{2}{ac}\:=\mathrm{0} \\ $$$${or}\:{bc}\:+{ab}\:=\mathrm{2}{ac} \\ $$$${or}\:\frac{{bc}}{{abc}}+\frac{{ab}}{{abc}}=\frac{\mathrm{2}{ac}}{{abc}}\: \\ $$$${or}\:\frac{\mathrm{1}}{{a}}+\frac{\mathrm{1}}{{c}}=\frac{\mathrm{2}}{{b}}\left({Proved}\right) \\ $$