Question-100158

Question Number 100158 by Algoritm last updated on 25/Jun/20

Answered by MJS last updated on 25/Jun/20

u=(√(x+3)) ⇔ x=u^2 −3∧u≥0 (u+4))(√(3u^2 −11))−3)=3u^2 −20 (u+4)(√(3u^2 −11))=3u^2 +3u−8 u^4 −u^3 −((38)/3)u^2 +((20)/3)u+40=0 trying factors of 40 ⇒ u_1 =3 ⇒ x_1 =6 u^3 +2u^2 −((20)/3)u−((40)/3)=0 trying ⇒ u_2 =−2 not possible because u≥0 u^2 −((20)/3)=0 ⇒ u_(3, 4) =±((2(√(15)))/3) ⇒ x_2 =((11)/3)

$${u}=\sqrt{{x}+\mathrm{3}}\:\Leftrightarrow\:{x}={u}^{\mathrm{2}} −\mathrm{3}\wedge{u}\geqslant\mathrm{0} \\ $$$$\left.\left(\left.{u}+\mathrm{4}\right)\right)\sqrt{\mathrm{3}{u}^{\mathrm{2}} −\mathrm{11}}−\mathrm{3}\right)=\mathrm{3}{u}^{\mathrm{2}} −\mathrm{20} \\ $$$$\left({u}+\mathrm{4}\right)\sqrt{\mathrm{3}{u}^{\mathrm{2}} −\mathrm{11}}=\mathrm{3}{u}^{\mathrm{2}} +\mathrm{3}{u}−\mathrm{8} \\ $$$${u}^{\mathrm{4}} −{u}^{\mathrm{3}} −\frac{\mathrm{38}}{\mathrm{3}}{u}^{\mathrm{2}} +\frac{\mathrm{20}}{\mathrm{3}}{u}+\mathrm{40}=\mathrm{0} \\ $$$$\mathrm{trying}\:\mathrm{factors}\:\mathrm{of}\:\mathrm{40} \\ $$$$\Rightarrow\:{u}_{\mathrm{1}} =\mathrm{3}\:\Rightarrow\:{x}_{\mathrm{1}} =\mathrm{6} \\ $$$${u}^{\mathrm{3}} +\mathrm{2}{u}^{\mathrm{2}} −\frac{\mathrm{20}}{\mathrm{3}}{u}−\frac{\mathrm{40}}{\mathrm{3}}=\mathrm{0} \\ $$$$\mathrm{trying}\:\Rightarrow\:{u}_{\mathrm{2}} =−\mathrm{2}\:\mathrm{not}\:\mathrm{possible}\:\mathrm{because}\:{u}\geqslant\mathrm{0} \\ $$$${u}^{\mathrm{2}} −\frac{\mathrm{20}}{\mathrm{3}}=\mathrm{0} \\ $$$$\Rightarrow\:{u}_{\mathrm{3},\:\mathrm{4}} =\pm\frac{\mathrm{2}\sqrt{\mathrm{15}}}{\mathrm{3}}\:\Rightarrow\:{x}_{\mathrm{2}} =\frac{\mathrm{11}}{\mathrm{3}} \\ $$

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Question-100158

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