Question and Answers Forum

All Questions      Topic List

Algebra Questions

Previous in All Question      Next in All Question      

Previous in Algebra      Next in Algebra      

Question Number 138150 by liberty last updated on 10/Apr/21

Given (x+(√(x^2 +1)))(y+(√(y^2 +4)))=9 find the value of x(√(y^2 +4)) + y(√(x^2 +1)) .

$${Given}\:\left({x}+\sqrt{{x}^{\mathrm{2}} +\mathrm{1}}\right)\left({y}+\sqrt{{y}^{\mathrm{2}} +\mathrm{4}}\right)=\mathrm{9} \\ $$$${find}\:{the}\:{value}\:{of}\: \\ $$$$\:\:\:\:\:\:\:{x}\sqrt{{y}^{\mathrm{2}} +\mathrm{4}}\:+\:{y}\sqrt{{x}^{\mathrm{2}} +\mathrm{1}}\:. \\ $$

Answered by EDWIN88 last updated on 10/Apr/21

set { ((x=cot 2α)),((y=2cot 2β)) :} with 0<α,β<π/2 we get determinant (((x+(√(x^2 +1)) = ((cos 2α)/(sin 2α))+(1/(sin 2α))=cot α)),((y+(√(y^2 +4)) = ((2cos 2β)/(sin 2β))+(2/(sin 2β)) = 2cot β))) then from the given equation 2cot α.cot β = a ((cos α cos β)/(sin α sin β)) = (a/2) we want to compute x(√(y^2 +4)) +y(√(x^2 +1)) = cot 2α ((2/(sin 2β)))+ 2cot 2β ((1/(sin 2α)))= ((2(cos 2α+cos 2β))/(sin 2α sin 2β)) = ((cos 2α +cos 2β)/(2(cos α cos β)(sin α sin β))) = ((2−2sin^2 α−2sin^2 β )/(2((a/2))sin^2 α sin^2 β)) =(2/a) (((1−sin^2 α−sin^2 β)/(sin^2 α sin^2 β))) = (2/a)((1+cot^2 α)(1+cot^2 β)−1−cot^2 α−1−cot^2 β) =(2/a)(cot^2 α cot^2 β−1) =(2/a)((a^2 /4)−1) = (2/a)(((a^2 −4)/4))= ((a^2 −4)/(2a)) in this case a=9 ; we get ((81−4)/(18))=((77)/(18))

$${set}\:\begin{cases}{{x}=\mathrm{cot}\:\mathrm{2}\alpha}\\{{y}=\mathrm{2cot}\:\mathrm{2}\beta}\end{cases}\:{with}\:\mathrm{0}<\alpha,\beta<\pi/\mathrm{2} \\ $$$${we}\:{get}\:\begin{array}{|c|c|}{{x}+\sqrt{{x}^{\mathrm{2}} +\mathrm{1}}\:=\:\frac{\mathrm{cos}\:\mathrm{2}\alpha}{\mathrm{sin}\:\mathrm{2}\alpha}+\frac{\mathrm{1}}{\mathrm{sin}\:\mathrm{2}\alpha}=\mathrm{cot}\:\alpha}\\{{y}+\sqrt{{y}^{\mathrm{2}} +\mathrm{4}}\:=\:\frac{\mathrm{2cos}\:\mathrm{2}\beta}{\mathrm{sin}\:\mathrm{2}\beta}+\frac{\mathrm{2}}{\mathrm{sin}\:\mathrm{2}\beta}\:=\:\mathrm{2cot}\:\beta}\\\hline\end{array} \\ $$$${then}\:{from}\:{the}\:{given}\:{equation} \\ $$$$\mathrm{2cot}\:\alpha.\mathrm{cot}\:\beta\:=\:{a}\: \:\frac{\mathrm{cos}\:\alpha\:\mathrm{cos}\:\beta}{\mathrm{sin}\:\alpha\:\mathrm{sin}\:\beta}\:=\:\frac{{a}}{\mathrm{2}} \\ $$$${we}\:{want}\:{to}\:{compute}\:{x}\sqrt{{y}^{\mathrm{2}} +\mathrm{4}}\:+{y}\sqrt{{x}^{\mathrm{2}} +\mathrm{1}}\:= \\ $$$$\mathrm{cot}\:\mathrm{2}\alpha\:\left(\frac{\mathrm{2}}{\mathrm{sin}\:\mathrm{2}\beta}\right)+\:\mathrm{2cot}\:\mathrm{2}\beta\:\left(\frac{\mathrm{1}}{\mathrm{sin}\:\mathrm{2}\alpha}\right)= \\ $$$$\frac{\mathrm{2}\left(\mathrm{cos}\:\mathrm{2}\alpha+\mathrm{cos}\:\mathrm{2}\beta\right)}{\mathrm{sin}\:\mathrm{2}\alpha\:\mathrm{sin}\:\mathrm{2}\beta}\:=\:\frac{\mathrm{cos}\:\mathrm{2}\alpha\:+\mathrm{cos}\:\mathrm{2}\beta}{\mathrm{2}\left(\mathrm{cos}\:\alpha\:\mathrm{cos}\:\beta\right)\left(\mathrm{sin}\:\alpha\:\mathrm{sin}\:\beta\right)} \\ $$$$=\:\frac{\mathrm{2}−\mathrm{2sin}^{\mathrm{2}} \alpha−\mathrm{2sin}\:^{\mathrm{2}} \beta\:}{\mathrm{2}\left(\frac{{a}}{\mathrm{2}}\right)\mathrm{sin}\:^{\mathrm{2}} \alpha\:\mathrm{sin}\:^{\mathrm{2}} \beta}\: \\ $$$$=\frac{\mathrm{2}}{{a}}\:\left(\frac{\mathrm{1}−\mathrm{sin}\:^{\mathrm{2}} \alpha−\mathrm{sin}\:^{\mathrm{2}} \beta}{\mathrm{sin}\:^{\mathrm{2}} \alpha\:\mathrm{sin}\:^{\mathrm{2}} \beta}\right) \\ $$$$=\:\frac{\mathrm{2}}{{a}}\left(\left(\mathrm{1}+\mathrm{cot}\:^{\mathrm{2}} \alpha\right)\left(\mathrm{1}+\mathrm{cot}\:^{\mathrm{2}} \beta\right)−\mathrm{1}−\mathrm{cot}\:^{\mathrm{2}} \alpha−\mathrm{1}−\mathrm{cot}\:^{\mathrm{2}} \beta\right) \\ $$$$=\frac{\mathrm{2}}{{a}}\left(\mathrm{cot}\:^{\mathrm{2}} \alpha\:\mathrm{cot}\:^{\mathrm{2}} \beta−\mathrm{1}\right) \\ $$$$=\frac{\mathrm{2}}{{a}}\left(\frac{{a}^{\mathrm{2}} }{\mathrm{4}}−\mathrm{1}\right)\:=\:\frac{\mathrm{2}}{{a}}\left(\frac{{a}^{\mathrm{2}} −\mathrm{4}}{\mathrm{4}}\right)=\:\frac{{a}^{\mathrm{2}} −\mathrm{4}}{\mathrm{2}{a}} \\ $$$${in}\:{this}\:{case}\:{a}=\mathrm{9}\:;\:{we}\:{get}\:\frac{\mathrm{81}−\mathrm{4}}{\mathrm{18}}=\frac{\mathrm{77}}{\mathrm{18}} \\ $$

Commented by liberty last updated on 11/Apr/21

nice

$${nice} \\ $$

Answered by MJS_new last updated on 10/Apr/21

u=x+(√(x^2 +1)) ⇔ x=((u^2 −1)/(2u)) v=y+(√(y^2 +4)) ⇔ y=((v^2 −4)/(2v)) v=(9/u) ⇒ y=((81−4u^2 )/(18u)) x(√(y^2 +4))+y(√(x^2 +1))=...=((77)/(18))

$${u}={x}+\sqrt{{x}^{\mathrm{2}} +\mathrm{1}}\:\Leftrightarrow\:{x}=\frac{{u}^{\mathrm{2}} −\mathrm{1}}{\mathrm{2}{u}} \\ $$$${v}={y}+\sqrt{{y}^{\mathrm{2}} +\mathrm{4}}\:\Leftrightarrow\:{y}=\frac{{v}^{\mathrm{2}} −\mathrm{4}}{\mathrm{2}{v}} \\ $$$${v}=\frac{\mathrm{9}}{{u}}\:\Rightarrow\:{y}=\frac{\mathrm{81}−\mathrm{4}{u}^{\mathrm{2}} }{\mathrm{18}{u}} \\ $$$${x}\sqrt{{y}^{\mathrm{2}} +\mathrm{4}}+{y}\sqrt{{x}^{\mathrm{2}} +\mathrm{1}}=...=\frac{\mathrm{77}}{\mathrm{18}} \\ $$

Commented by liberty last updated on 11/Apr/21

short cut?

$${short}\:{cut}? \\ $$

Commented by MJS_new last updated on 11/Apr/21

well I guess it′s not necessary to type all the algebraic steps... btw. we can also solve the given equation for y: X(y+(√(y^2 +4)))=9 ⇒ y=((81−4X^2 )/(18X))=((−85x+77(√(x^2 +1)))/(18)) ⇒ x(√(y^2 +4))+y(√(x^2 +1))=...=((77)/(18))

$$\mathrm{well}\:\mathrm{I}\:\mathrm{guess}\:\mathrm{it}'\mathrm{s}\:\mathrm{not}\:\mathrm{necessary}\:\mathrm{to}\:\mathrm{type}\:\mathrm{all}\:\mathrm{the} \\ $$$$\mathrm{algebraic}\:\mathrm{steps}... \\ $$$$\mathrm{btw}.\:\mathrm{we}\:\mathrm{can}\:\mathrm{also}\:\mathrm{solve}\:\mathrm{the}\:\mathrm{given}\:\mathrm{equation} \\ $$$$\mathrm{for}\:{y}: \\ $$$${X}\left({y}+\sqrt{{y}^{\mathrm{2}} +\mathrm{4}}\right)=\mathrm{9}\:\Rightarrow\:{y}=\frac{\mathrm{81}−\mathrm{4}{X}^{\mathrm{2}} }{\mathrm{18}{X}}=\frac{−\mathrm{85}{x}+\mathrm{77}\sqrt{{x}^{\mathrm{2}} +\mathrm{1}}}{\mathrm{18}} \\ $$$$\Rightarrow\:{x}\sqrt{{y}^{\mathrm{2}} +\mathrm{4}}+{y}\sqrt{{x}^{\mathrm{2}} +\mathrm{1}}=...=\frac{\mathrm{77}}{\mathrm{18}} \\ $$

Commented by SLVR last updated on 11/Apr/21

this part of solution is marvoles

$${this}\:{part}\:{of}\:{solution}\:{is}\:{marvoles} \\ $$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com