Question-146315

Question Number 146315 by mnjuly1970 last updated on 12/Jul/21

Answered by Olaf_Thorendsen last updated on 13/Jul/21

(√((x(x+5))/(x+2)))+(√(x+3)) = ((√(16x+32))/(x+2)) (√((x(x+5))/(x+2)))+(√(x+3)) = (4/( (√(x+2)))) (√(x(x+5)))+(√((x+3)(x+2))) = 4 Let u = x+(5/2) (√((u−(5/2))(u+(5/2))))+(√((u+(1/2))(u−(1/2)))) = 4 (√(u^2 −((25)/4)))+(√(u^2 −(1/4))) = 4 Let v = u^2 −(1/4) (√(v−6)) = 4−(√v) ⇒ v−6 = 16−8(√v)+v (√v) = ((11)/4) ⇒ v = ((121)/(16)) u^2 = v+(1/4) = ((125)/(16)) ⇒ u = ±((5(√5))/4) x = u−(5/2) = −(5/4)(2±(√5)) x = −(5/4)(2+(√5)) ≈ −5,29 (impossible because x+2 > 0) x = −(5/4)(2−(√5)) ≈ +0,295 OK Finally x = −(5/4)(2−(√5))

$$\sqrt{\frac{{x}\left({x}+\mathrm{5}\right)}{{x}+\mathrm{2}}}+\sqrt{{x}+\mathrm{3}}\:=\:\frac{\sqrt{\mathrm{16}{x}+\mathrm{32}}}{{x}+\mathrm{2}} \\ $$$$\sqrt{\frac{{x}\left({x}+\mathrm{5}\right)}{{x}+\mathrm{2}}}+\sqrt{{x}+\mathrm{3}}\:=\:\frac{\mathrm{4}}{\:\sqrt{{x}+\mathrm{2}}} \\ $$$$\sqrt{{x}\left({x}+\mathrm{5}\right)}+\sqrt{\left({x}+\mathrm{3}\right)\left({x}+\mathrm{2}\right)}\:=\:\mathrm{4} \\ $$$$\mathrm{Let}\:{u}\:=\:{x}+\frac{\mathrm{5}}{\mathrm{2}} \\ $$$$\sqrt{\left({u}−\frac{\mathrm{5}}{\mathrm{2}}\right)\left({u}+\frac{\mathrm{5}}{\mathrm{2}}\right)}+\sqrt{\left({u}+\frac{\mathrm{1}}{\mathrm{2}}\right)\left({u}−\frac{\mathrm{1}}{\mathrm{2}}\right)}\:=\:\mathrm{4} \\ $$$$\sqrt{{u}^{\mathrm{2}} −\frac{\mathrm{25}}{\mathrm{4}}}+\sqrt{{u}^{\mathrm{2}} −\frac{\mathrm{1}}{\mathrm{4}}}\:=\:\mathrm{4} \\ $$$$\mathrm{Let}\:{v}\:=\:{u}^{\mathrm{2}} −\frac{\mathrm{1}}{\mathrm{4}} \\ $$$$\sqrt{{v}−\mathrm{6}}\:=\:\mathrm{4}−\sqrt{{v}} \\ $$$$\Rightarrow\:{v}−\mathrm{6}\:=\:\mathrm{16}−\mathrm{8}\sqrt{{v}}+{v} \\ $$$$\sqrt{{v}}\:=\:\frac{\mathrm{11}}{\mathrm{4}}\:\Rightarrow\:{v}\:=\:\frac{\mathrm{121}}{\mathrm{16}} \\ $$$${u}^{\mathrm{2}} \:=\:{v}+\frac{\mathrm{1}}{\mathrm{4}}\:=\:\frac{\mathrm{125}}{\mathrm{16}}\:\Rightarrow\:{u}\:=\:\pm\frac{\mathrm{5}\sqrt{\mathrm{5}}}{\mathrm{4}} \\ $$$${x}\:=\:{u}−\frac{\mathrm{5}}{\mathrm{2}}\:=\:−\frac{\mathrm{5}}{\mathrm{4}}\left(\mathrm{2}\pm\sqrt{\mathrm{5}}\right) \\ $$$${x}\:=\:−\frac{\mathrm{5}}{\mathrm{4}}\left(\mathrm{2}+\sqrt{\mathrm{5}}\right)\:\approx\:−\mathrm{5},\mathrm{29} \\ $$$$\left(\mathrm{impossible}\:\mathrm{because}\:{x}+\mathrm{2}\:>\:\mathrm{0}\right) \\ $$$${x}\:=\:−\frac{\mathrm{5}}{\mathrm{4}}\left(\mathrm{2}−\sqrt{\mathrm{5}}\right)\:\approx\:+\mathrm{0},\mathrm{295}\:\mathrm{OK} \\ $$$$ \\ $$$$\mathrm{Finally}\:{x}\:=\:−\frac{\mathrm{5}}{\mathrm{4}}\left(\mathrm{2}−\sqrt{\mathrm{5}}\right) \\ $$

Commented by mnjuly1970 last updated on 13/Jul/21

$${thank}\:{you}\:{mr}\:{olaf} \\ $$

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