Question Number 204521 by explorerapollo last updated on 20/Feb/24
Commented by TonyCWX08 last updated on 20/Feb/24
$${I}\:{don}'{t}\:{understand}\:{your}\:{request}\:{here}. \\ $$
Answered by mr W last updated on 20/Feb/24
Commented by mr W last updated on 20/Feb/24
$$\mathrm{2}{C}=\left({a}−\frac{\mathrm{2}{B}}{{b}}\right)\left({b}−\frac{\mathrm{2}{A}}{{a}}\right)={ab}−\mathrm{2}{A}−\mathrm{2}{B}+\frac{\mathrm{4}{AB}}{{ab}} \\ $$$${ab}^{\mathrm{2}} −\mathrm{2}\left({A}+{B}+{C}\right){ab}+\mathrm{4}{AB}=\mathrm{0} \\ $$$$\Rightarrow{ab}={A}+{B}+{C}+\sqrt{\left({A}+{B}+{C}\right)^{\mathrm{2}} −\mathrm{4}{AB}} \\ $$$$ \\ $$$${X}={ab}−{A}−{B}−{C}=\sqrt{\left({A}+{B}+{C}\right)^{\mathrm{2}} −\mathrm{4}{AB}} \\ $$$${with}\:{A}=\mathrm{6},\:{B}=\mathrm{2},\:{C}=\mathrm{5} \\ $$$$\Rightarrow{X}=\sqrt{\left(\mathrm{6}+\mathrm{2}+\mathrm{5}\right)^{\mathrm{2}} −\mathrm{4}×\mathrm{6}×\mathrm{2}}=\mathrm{11} \\ $$