Question and Answers Forum

All Questions      Topic List

Vector Questions

Previous in All Question      Next in All Question      

Previous in Vector      Next in Vector      

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

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

Terms of Service

Privacy Policy

Contact: info@tinkutara.com