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Question Number 70898 by Henri Boucatchou last updated on 09/Oct/19

Solve : 1.) (√(x−2)) + (√(4−x)) = x^2 −6x+11 2.) x^4 + x^3 − 2ax^2 − ax + a^2 = 0

$$\boldsymbol{{Solve}}\::\: \\ $$$$\left.\mathrm{1}.\right)\:\:\sqrt{\boldsymbol{{x}}−\mathrm{2}}\:+\:\sqrt{\mathrm{4}−\boldsymbol{{x}}}\:=\:\boldsymbol{{x}}^{\mathrm{2}} −\mathrm{6}\boldsymbol{{x}}+\mathrm{11} \\ $$$$\left.\mathrm{2}.\right)\:\:\boldsymbol{{x}}^{\mathrm{4}} \:+\:\boldsymbol{{x}}^{\mathrm{3}} \:−\:\mathrm{2}\boldsymbol{{ax}}^{\mathrm{2}} \:−\:\boldsymbol{{ax}}\:+\:\boldsymbol{{a}}^{\mathrm{2}} \:=\:\mathrm{0} \\ $$

Answered by MJS last updated on 09/Oct/19

2) x^4 +x^3 −2ax^2 −ax+a^2 =0 (x^2 −a)(x^2 +x−a)=0 ⇒ x=±(√a) ∨ x=−(1/2)±((√(4a+1))/2)

$$\left.\mathrm{2}\right) \\ $$$${x}^{\mathrm{4}} +{x}^{\mathrm{3}} −\mathrm{2}{ax}^{\mathrm{2}} −{ax}+{a}^{\mathrm{2}} =\mathrm{0} \\ $$$$\left({x}^{\mathrm{2}} −{a}\right)\left({x}^{\mathrm{2}} +{x}−{a}\right)=\mathrm{0} \\ $$$$\Rightarrow\:{x}=\pm\sqrt{{a}}\:\vee\:{x}=−\frac{\mathrm{1}}{\mathrm{2}}\pm\frac{\sqrt{\mathrm{4}{a}+\mathrm{1}}}{\mathrm{2}} \\ $$

Answered by MJS last updated on 09/Oct/19

1) (√(x−2))+(√(4−x))=x^2 −6x+11 t=x−3 ⇔ x=t+3 (√(1−t))+(√(1+t))=t^2 +2 squaring 2+2(√(t^2 −1))=t^4 +4t^2 +4 2(√(1−t^2 ))=t^4 +4t^2 +2 squaring 4−4t^2 =8t^6 +20t^4 +16t^2 +4 t^2 (t^6 +8t^4 +20t^2 +20)=0 t_1 =t_2 =0 no other real solution ⇒ x=3

$$\left.\mathrm{1}\right) \\ $$$$\sqrt{{x}−\mathrm{2}}+\sqrt{\mathrm{4}−{x}}={x}^{\mathrm{2}} −\mathrm{6}{x}+\mathrm{11} \\ $$$${t}={x}−\mathrm{3}\:\Leftrightarrow\:{x}={t}+\mathrm{3} \\ $$$$\sqrt{\mathrm{1}−{t}}+\sqrt{\mathrm{1}+{t}}={t}^{\mathrm{2}} +\mathrm{2} \\ $$$$\mathrm{squaring} \\ $$$$\mathrm{2}+\mathrm{2}\sqrt{{t}^{\mathrm{2}} −\mathrm{1}}={t}^{\mathrm{4}} +\mathrm{4}{t}^{\mathrm{2}} +\mathrm{4} \\ $$$$\mathrm{2}\sqrt{\mathrm{1}−{t}^{\mathrm{2}} }={t}^{\mathrm{4}} +\mathrm{4}{t}^{\mathrm{2}} +\mathrm{2} \\ $$$$\mathrm{squaring} \\ $$$$\mathrm{4}−\mathrm{4}{t}^{\mathrm{2}} =\mathrm{8}{t}^{\mathrm{6}} +\mathrm{20}{t}^{\mathrm{4}} +\mathrm{16}{t}^{\mathrm{2}} +\mathrm{4} \\ $$$${t}^{\mathrm{2}} \left({t}^{\mathrm{6}} +\mathrm{8}{t}^{\mathrm{4}} +\mathrm{20}{t}^{\mathrm{2}} +\mathrm{20}\right)=\mathrm{0} \\ $$$${t}_{\mathrm{1}} ={t}_{\mathrm{2}} =\mathrm{0} \\ $$$$\mathrm{no}\:\mathrm{other}\:\mathrm{real}\:\mathrm{solution} \\ $$$$\Rightarrow\:{x}=\mathrm{3} \\ $$

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