Question Number 70898 by Henri Boucatchou last updated on 09/Oct/19
$$\boldsymbol{{Solve}}\::\: \\ $$$$\left.\mathrm{1}.\right)\:\:\sqrt{\boldsymbol{{x}}−\mathrm{2}}\:+\:\sqrt{\mathrm{4}−\boldsymbol{{x}}}\:=\:\boldsymbol{{x}}^{\mathrm{2}} −\mathrm{6}\boldsymbol{{x}}+\mathrm{11} \\ $$$$\left.\mathrm{2}.\right)\:\:\boldsymbol{{x}}^{\mathrm{4}} \:+\:\boldsymbol{{x}}^{\mathrm{3}} \:−\:\mathrm{2}\boldsymbol{{ax}}^{\mathrm{2}} \:−\:\boldsymbol{{ax}}\:+\:\boldsymbol{{a}}^{\mathrm{2}} \:=\:\mathrm{0} \\ $$
Answered by MJS last updated on 09/Oct/19
$$\left.\mathrm{2}\right) \\ $$$${x}^{\mathrm{4}} +{x}^{\mathrm{3}} −\mathrm{2}{ax}^{\mathrm{2}} −{ax}+{a}^{\mathrm{2}} =\mathrm{0} \\ $$$$\left({x}^{\mathrm{2}} −{a}\right)\left({x}^{\mathrm{2}} +{x}−{a}\right)=\mathrm{0} \\ $$$$\Rightarrow\:{x}=\pm\sqrt{{a}}\:\vee\:{x}=−\frac{\mathrm{1}}{\mathrm{2}}\pm\frac{\sqrt{\mathrm{4}{a}+\mathrm{1}}}{\mathrm{2}} \\ $$
Answered by MJS last updated on 09/Oct/19
$$\left.\mathrm{1}\right) \\ $$$$\sqrt{{x}−\mathrm{2}}+\sqrt{\mathrm{4}−{x}}={x}^{\mathrm{2}} −\mathrm{6}{x}+\mathrm{11} \\ $$$${t}={x}−\mathrm{3}\:\Leftrightarrow\:{x}={t}+\mathrm{3} \\ $$$$\sqrt{\mathrm{1}−{t}}+\sqrt{\mathrm{1}+{t}}={t}^{\mathrm{2}} +\mathrm{2} \\ $$$$\mathrm{squaring} \\ $$$$\mathrm{2}+\mathrm{2}\sqrt{{t}^{\mathrm{2}} −\mathrm{1}}={t}^{\mathrm{4}} +\mathrm{4}{t}^{\mathrm{2}} +\mathrm{4} \\ $$$$\mathrm{2}\sqrt{\mathrm{1}−{t}^{\mathrm{2}} }={t}^{\mathrm{4}} +\mathrm{4}{t}^{\mathrm{2}} +\mathrm{2} \\ $$$$\mathrm{squaring} \\ $$$$\mathrm{4}−\mathrm{4}{t}^{\mathrm{2}} =\mathrm{8}{t}^{\mathrm{6}} +\mathrm{20}{t}^{\mathrm{4}} +\mathrm{16}{t}^{\mathrm{2}} +\mathrm{4} \\ $$$${t}^{\mathrm{2}} \left({t}^{\mathrm{6}} +\mathrm{8}{t}^{\mathrm{4}} +\mathrm{20}{t}^{\mathrm{2}} +\mathrm{20}\right)=\mathrm{0} \\ $$$${t}_{\mathrm{1}} ={t}_{\mathrm{2}} =\mathrm{0} \\ $$$$\mathrm{no}\:\mathrm{other}\:\mathrm{real}\:\mathrm{solution} \\ $$$$\Rightarrow\:{x}=\mathrm{3} \\ $$