Question Number 224028 by behi834171 last updated on 15/Aug/25
Answered by Rasheed.Sindhi last updated on 15/Aug/25
$$\frac{\left({x}+{c}^{\mathrm{2}} \right)\left({x}−{ab}\right)}{\left({x}+{a}^{\mathrm{2}} \right)\left({x}+{b}^{\mathrm{2}} \right)}=\frac{{c}}{{a}+{b}} \\ $$$$\left({a}+{b}\right)\left({x}^{\mathrm{2}} +\left(−{ab}+{c}^{\mathrm{2}} \right){x}−{abc}^{\mathrm{2}} \right) \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:={c}\left({x}^{\mathrm{2}} +\left({a}^{\mathrm{2}} +{b}^{\mathrm{2}} \right){x}+{a}^{\mathrm{2}} {b}^{\mathrm{2}} \right) \\ $$$$\left({a}+{b}−{c}\right){x}^{\mathrm{2}} +\left(−{ab}+{c}^{\mathrm{2}} −{a}^{\mathrm{2}} −{b}^{\mathrm{2}} \right){x}−{abc}^{\mathrm{2}} −{a}^{\mathrm{2}} {b}^{\mathrm{2}} =\mathrm{0} \\ $$$$\left({a}+{b}−{c}\right){x}^{\mathrm{2}} −\left({a}^{\mathrm{2}} +{b}^{\mathrm{2}} −{c}^{\mathrm{2}} +{ab}\right){x}−{ab}\left({c}^{\mathrm{2}} −{ab}\right)=\mathrm{0} \\ $$$${x}=\frac{{a}^{\mathrm{2}} +{b}^{\mathrm{2}} −{c}^{\mathrm{2}} +{ab}\:\pm\sqrt{\left({a}^{\mathrm{2}} +{b}^{\mathrm{2}} −{c}^{\mathrm{2}} +{ab}\right)^{\mathrm{2}} −\mathrm{4}\left({a}+{b}−{c}\right)\left(−{ab}\left({c}^{\mathrm{2}} −{ab}\right)\right)}}{\mathrm{2}\left({a}+{b}−{c}\right)} \\ $$$${x}=\frac{{a}^{\mathrm{2}} +{b}^{\mathrm{2}} −{c}^{\mathrm{2}} +{ab}\:\pm\sqrt{\left({a}^{\mathrm{2}} +{b}^{\mathrm{2}} −{c}^{\mathrm{2}} +{ab}\right)^{\mathrm{2}} +\mathrm{4}{ab}\left({a}+{b}−{c}\right)\left({c}^{\mathrm{2}} −{ab}\right)}}{\mathrm{2}\left({a}+{b}−{c}\right)} \\ $$$${Simplifying}\:{above}\:{is}\:{very}\:{hard}. \\ $$$$\: \\ $$
Commented by Ghisom last updated on 15/Aug/25
$$\mathrm{we}\:\mathrm{can}\:\mathrm{factorize}\:\Delta \\ $$$$\Rightarrow \\ $$$${x}=−\frac{{abc}}{{a}+{b}−{c}}\vee{x}={ab}+{ac}+{bc} \\ $$
Commented by Rasheed.Sindhi last updated on 16/Aug/25
$${Thanks}\:{Sir}! \\ $$