Question-162382

Question Number 162382 by HongKing last updated on 29/Dec/21

Answered by Rasheed.Sindhi last updated on 29/Dec/21

{: (( (x+(√(x^2 +1)) )(y+(√(y^2 +1)) )=2011)),(( x+y=((2010)/( (√(2011)))) )) } _(−) (x+(√(x^2 +1)) )∙(((y+(√(y^2 +1)) )(y−(√(y^2 +1)) ))/((y−(√(y^2 +1)) )))=2011 −((x+(√(x^2 +1)) )/(y−(√(y^2 +1)) ))=2011 ((x+(√(x^2 +1)) )/(y−(√(y^2 +1)) ))+1=−2011+1 ((x+y)/(y−(√(y^2 +1)) ))=−2010 (((2010)/( (√(2011)) ))/(y−(√(y^2 +1)) ))=−2010 (1/( (√(2011)) (y−(√(y^2 +1)) )))=−1 y−(√(y^2 +1)) =−(1/( (√(2011)))) 2y^2 +1−(1/(2011))=2y(√(y^2 +1)) y^2 +((1005)/(2011))=y(√(y^2 +1)) 2011y^2 +1005=2011y(√(y^2 +1)) 2011^2 y^4 ^(×) +2∙2011∙1005y^2 +1005^2 =2011^2 y^2 (y^2 +1) =2011^2 y^4 ^(×) +2011^2 y^2 (2011∙2010−2011^2 )y^2 =−1005^2 y^2 =((−1005^2 )/(2011(2010−2011)))=((1005^2 )/(2011)) y=±((1005)/( (√(2011)))) As the equations are symmetric in x & y So x=±((1005)/( (√(2011)))) (−((1005)/( (√(2011)))),−((1005)/( (√(2011))))) doesn′t satisfy x+y=((2010)/( (√(2011)))) Hence (x,y)=(((1005)/( (√(2011)))) , ((1005)/( (√(2011)))))

$$\underset{−} {\left.\begin{matrix}{\:\:\:\:\:\:\:\:\left(\mathrm{x}+\sqrt{\mathrm{x}^{\mathrm{2}} +\mathrm{1}}\:\right)\left(\mathrm{y}+\sqrt{\mathrm{y}^{\mathrm{2}} +\mathrm{1}}\:\right)=\mathrm{2011}}\\{\:\:\:\:\:\:\:\:\:\mathrm{x}+\mathrm{y}=\frac{\mathrm{2010}}{\:\sqrt{\mathrm{2011}}}\:\:}\end{matrix}\right\}\:\:\:\:\:\:\:\:\:\:}\:\: \\ $$$$\left(\mathrm{x}+\sqrt{\mathrm{x}^{\mathrm{2}} +\mathrm{1}}\:\right)\centerdot\frac{\left(\mathrm{y}+\sqrt{\mathrm{y}^{\mathrm{2}} +\mathrm{1}}\:\right)\left(\mathrm{y}−\sqrt{\mathrm{y}^{\mathrm{2}} +\mathrm{1}}\:\right)}{\left(\mathrm{y}−\sqrt{\mathrm{y}^{\mathrm{2}} +\mathrm{1}}\:\right)}=\mathrm{2011} \\ $$$$−\frac{\mathrm{x}+\sqrt{\mathrm{x}^{\mathrm{2}} +\mathrm{1}}\:}{\mathrm{y}−\sqrt{\mathrm{y}^{\mathrm{2}} +\mathrm{1}}\:}=\mathrm{2011} \\ $$$$\frac{\mathrm{x}+\sqrt{\mathrm{x}^{\mathrm{2}} +\mathrm{1}}\:}{\mathrm{y}−\sqrt{\mathrm{y}^{\mathrm{2}} +\mathrm{1}}\:}+\mathrm{1}=−\mathrm{2011}+\mathrm{1} \\ $$$$\frac{\mathrm{x}+\mathrm{y}}{\mathrm{y}−\sqrt{\mathrm{y}^{\mathrm{2}} +\mathrm{1}}\:}=−\mathrm{2010} \\ $$$$\frac{\frac{\mathrm{2010}}{\:\sqrt{\mathrm{2011}}\:}}{\mathrm{y}−\sqrt{\mathrm{y}^{\mathrm{2}} +\mathrm{1}}\:}=−\mathrm{2010} \\ $$$$\frac{\mathrm{1}}{\:\sqrt{\mathrm{2011}}\:\left(\mathrm{y}−\sqrt{\mathrm{y}^{\mathrm{2}} +\mathrm{1}}\:\right)}=−\mathrm{1} \\ $$$$\:\mathrm{y}−\sqrt{\mathrm{y}^{\mathrm{2}} +\mathrm{1}}\:=−\frac{\mathrm{1}}{\:\sqrt{\mathrm{2011}}} \\ $$$$\mathrm{2y}^{\mathrm{2}} +\mathrm{1}−\frac{\mathrm{1}}{\mathrm{2011}}=\mathrm{2y}\sqrt{\mathrm{y}^{\mathrm{2}} +\mathrm{1}}\: \\ $$$$\mathrm{y}^{\mathrm{2}} +\frac{\mathrm{1005}}{\mathrm{2011}}=\mathrm{y}\sqrt{\mathrm{y}^{\mathrm{2}} +\mathrm{1}}\: \\ $$$$\mathrm{2011y}^{\mathrm{2}} +\mathrm{1005}=\mathrm{2011y}\sqrt{\mathrm{y}^{\mathrm{2}} +\mathrm{1}}\: \\ $$$$\overset{×} {\mathrm{2011}^{\mathrm{2}} \mathrm{y}^{\mathrm{4}} }+\mathrm{2}\centerdot\mathrm{2011}\centerdot\mathrm{1005y}^{\mathrm{2}} +\mathrm{1005}^{\mathrm{2}} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\mathrm{2011}^{\mathrm{2}} \mathrm{y}^{\mathrm{2}} \left(\mathrm{y}^{\mathrm{2}} +\mathrm{1}\right) \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\overset{×} {\mathrm{2011}^{\mathrm{2}} \mathrm{y}^{\mathrm{4}} }+\mathrm{2011}^{\mathrm{2}} \mathrm{y}^{\mathrm{2}} \\ $$$$\left(\mathrm{2011}\centerdot\mathrm{2010}−\mathrm{2011}^{\mathrm{2}} \right)\mathrm{y}^{\mathrm{2}} =−\mathrm{1005}^{\mathrm{2}} \\ $$$$\mathrm{y}^{\mathrm{2}} =\frac{−\mathrm{1005}^{\mathrm{2}} }{\mathrm{2011}\left(\mathrm{2010}−\mathrm{2011}\right)}=\frac{\mathrm{1005}^{\mathrm{2}} }{\mathrm{2011}} \\ $$$$\mathrm{y}=\pm\frac{\mathrm{1005}}{\:\sqrt{\mathrm{2011}}} \\ $$$$\mathrm{As}\:\mathrm{the}\:\mathrm{equations}\:\mathrm{are}\:\mathrm{symmetric}\:\mathrm{in}\:\mathrm{x}\:\&\:\mathrm{y} \\ $$$$\mathrm{So}\: \\ $$$$\mathrm{x}=\pm\frac{\mathrm{1005}}{\:\sqrt{\mathrm{2011}}} \\ $$$$\left(−\frac{\mathrm{1005}}{\:\sqrt{\mathrm{2011}}},−\frac{\mathrm{1005}}{\:\sqrt{\mathrm{2011}}}\right)\:\mathrm{doesn}'\mathrm{t}\:\mathrm{satisfy}\:\mathrm{x}+\mathrm{y}=\frac{\mathrm{2010}}{\:\sqrt{\mathrm{2011}}} \\ $$$$\mathrm{Hence} \\ $$$$\left(\mathrm{x},\mathrm{y}\right)=\left(\frac{\mathrm{1005}}{\:\sqrt{\mathrm{2011}}}\:,\:\frac{\mathrm{1005}}{\:\sqrt{\mathrm{2011}}}\right) \\ $$

Commented by HongKing last updated on 29/Dec/21

$$\mathrm{cool}\:\mathrm{my}\:\mathrm{dear}\:\mathrm{Sir}\:\mathrm{thank}\:\mathrm{you}\:\mathrm{so}\:\mathrm{much} \\ $$

Commented by mr W last updated on 29/Dec/21

how did you get this step sir? from ((x+(√(x^2 +1)) )/(y−(√(y^2 +1)) ))+1=−2011+1 to ((x+y)/(y−(√(y^2 +1)) ))=−2010

$${how}\:{did}\:{you}\:{get}\:{this}\:{step}\:{sir}? \\ $$$${from}\:\frac{\mathrm{x}+\sqrt{\mathrm{x}^{\mathrm{2}} +\mathrm{1}}\:}{\mathrm{y}−\sqrt{\mathrm{y}^{\mathrm{2}} +\mathrm{1}}\:}+\mathrm{1}=−\mathrm{2011}+\mathrm{1} \\ $$$${to}\:\frac{\mathrm{x}+\mathrm{y}}{\mathrm{y}−\sqrt{\mathrm{y}^{\mathrm{2}} +\mathrm{1}}\:}=−\mathrm{2010} \\ $$

Commented by mr W last updated on 29/Dec/21

for more generality can we change the question to { (((x+(√(x^2 +1)))(y+(√(y^2 +1)))=10)),((x+y=4)) :}

$${for}\:{more}\:{generality}\:{can}\:{we}\:{change} \\ $$$${the}\:{question}\:{to} \\ $$$$\begin{cases}{\left({x}+\sqrt{{x}^{\mathrm{2}} +\mathrm{1}}\right)\left({y}+\sqrt{{y}^{\mathrm{2}} +\mathrm{1}}\right)=\mathrm{10}}\\{{x}+{y}=\mathrm{4}}\end{cases} \\ $$

Commented by Rasheed.Sindhi last updated on 29/Dec/21

Mistake sir! I considered (√(x^2 +1)) & (√(y^2 +1)) same! I was very happy for eliminating x, but this is not so easy. 😭😭😭

$$\mathrm{Mistake}\:\boldsymbol{\mathrm{sir}}!\:\mathrm{I}\:\mathrm{considered}\:\:\sqrt{\mathrm{x}^{\mathrm{2}} +\mathrm{1}}\:\&\:\sqrt{\mathrm{y}^{\mathrm{2}} +\mathrm{1}} \\ $$$$\boldsymbol{\mathrm{same}}!\:\mathrm{I}\:\mathrm{was}\:\mathrm{very}\:\mathrm{happy}\:\mathrm{for}\:\mathrm{eliminating}\:\mathrm{x}, \\ $$$$\mathrm{but}\:\mathrm{this}\:\mathrm{is}\:\mathrm{not}\:\mathrm{so}\:\mathrm{easy}.\: \\ $$😭😭😭

Commented by mr W last updated on 29/Dec/21

$${thanks}\:{for}\:{replying}\:{sir}. \\ $$$${it}'{s}\:{really}\:{not}\:{easy}\:{to}\:{separate}\:{x},\:{y} \\ $$$${from}\:{each}\:{other}. \\ $$

Commented by Rasheed.Sindhi last updated on 29/Dec/21

I wonder sir,that HongKing sir also hasn′t noticed it! Perhaps the answer is accidently correct or he hadn′t answer of the question! Anyway thanks for your keen observation!

$${I}\:{wonder}\:\boldsymbol{{sir}},{that}\:{HongKing}\:{sir} \\ $$$${also}\:{hasn}'{t}\:{noticed}\:{it}!\:{Perhaps}\:{the} \\ $$$${answer}\:{is}\:{accidently}\:{correct}\:{or}\:{he} \\ $$$${hadn}'{t}\:{answer}\:{of}\:{the}\:{question}! \\ $$$${Anyway}\:{thanks}\:{for}\:{your}\:{keen}\:{observation}! \\ $$

Commented by mr W last updated on 29/Dec/21

$${the}\:{question}\:{is}\:{not}\:{general}\:{enough}. \\ $$$${both}\:{symmetric}\:{curves}\:{tangent}\:{each} \\ $$$${other},\:{so}\:{x}={y}\:{fulfills}\:{the}\:{solution}. \\ $$

Commented by Rasheed.Sindhi last updated on 29/Dec/21

Exacly sir, and this is the reason I reached correct answer by considering (√(x^2 +1)) =(√(y^2 +1)) ! Thanks again sir!

$$\mathcal{E}{xacly}\:{sir},\:{and}\:{this}\:{is}\:{the}\:{reason} \\ $$$${I}\:{reached}\:{correct}\:{answer}\:{by} \\ $$$${considering}\:\sqrt{{x}^{\mathrm{2}} +\mathrm{1}}\:=\sqrt{{y}^{\mathrm{2}} +\mathrm{1}}\:! \\ $$$$\mathcal{T}{hanks}\:{again}\:{sir}! \\ $$

Commented by mr W last updated on 29/Dec/21

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Commented by Ar Brandon last updated on 29/Dec/21

It′s really not easy to separate x and y from each other. They are deeply in love 🤗🤭

$$\mathrm{It}'\mathrm{s}\:\mathrm{really}\:\mathrm{not}\:\mathrm{easy}\:\mathrm{to}\:\mathrm{separate}\:{x}\:\mathrm{and}\:{y}\: \\ $$$$\mathrm{from}\:\mathrm{each}\:\mathrm{other}.\:\mathrm{They}\:\mathrm{are}\:\mathrm{deeply}\:\mathrm{in}\:\mathrm{love} \\ $$🤗🤭

Answered by MJS_new last updated on 29/Dec/21

(x+(√(x^2 +1)))(y+(√(y^2 +1)))=a x+y=b let u=x+(√(x^2 +1))∧v=y+(√(y^2 +1)) uv=a (1/2)(u−(1/u)+v−(1/v))=b ⇔ uv=a u+v=((2ab)/(a−1)) ⇔ v=(a/u) u^2 −((2ab)/(a−1))u+a=0 ⇔ u=((ab±(√(−a(a^2 −(b^2 +2)a+1))))/(a−1)) v=((ab∓(√(−a(a^2 −(b^2 +2)a+1))))/(a−1)) ⇒ x=(b/2)±(((a+1))/(2a(a−1)))(√(−a(a^2 −(b^2 +2)a+1))) y=(b/2)∓(((a+1))/(2a(a−1)))(√(−a(a^2 −(b^2 +2)a+1)))

$$\left({x}+\sqrt{{x}^{\mathrm{2}} +\mathrm{1}}\right)\left({y}+\sqrt{{y}^{\mathrm{2}} +\mathrm{1}}\right)={a} \\ $$$${x}+{y}={b} \\ $$$$\mathrm{let}\:{u}={x}+\sqrt{{x}^{\mathrm{2}} +\mathrm{1}}\wedge{v}={y}+\sqrt{{y}^{\mathrm{2}} +\mathrm{1}} \\ $$$${uv}={a} \\ $$$$\frac{\mathrm{1}}{\mathrm{2}}\left({u}−\frac{\mathrm{1}}{{u}}+{v}−\frac{\mathrm{1}}{{v}}\right)={b} \\ $$$$\Leftrightarrow \\ $$$${uv}={a} \\ $$$${u}+{v}=\frac{\mathrm{2}{ab}}{{a}−\mathrm{1}} \\ $$$$\Leftrightarrow \\ $$$${v}=\frac{{a}}{{u}} \\ $$$${u}^{\mathrm{2}} −\frac{\mathrm{2}{ab}}{{a}−\mathrm{1}}{u}+{a}=\mathrm{0} \\ $$$$\Leftrightarrow \\ $$$${u}=\frac{{ab}\pm\sqrt{−{a}\left({a}^{\mathrm{2}} −\left({b}^{\mathrm{2}} +\mathrm{2}\right){a}+\mathrm{1}\right)}}{{a}−\mathrm{1}} \\ $$$${v}=\frac{{ab}\mp\sqrt{−{a}\left({a}^{\mathrm{2}} −\left({b}^{\mathrm{2}} +\mathrm{2}\right){a}+\mathrm{1}\right)}}{{a}−\mathrm{1}} \\ $$$$\Rightarrow \\ $$$${x}=\frac{{b}}{\mathrm{2}}\pm\frac{\left({a}+\mathrm{1}\right)}{\mathrm{2}{a}\left({a}−\mathrm{1}\right)}\sqrt{−{a}\left({a}^{\mathrm{2}} −\left({b}^{\mathrm{2}} +\mathrm{2}\right){a}+\mathrm{1}\right)} \\ $$$${y}=\frac{{b}}{\mathrm{2}}\mp\frac{\left({a}+\mathrm{1}\right)}{\mathrm{2}{a}\left({a}−\mathrm{1}\right)}\sqrt{−{a}\left({a}^{\mathrm{2}} −\left({b}^{\mathrm{2}} +\mathrm{2}\right){a}+\mathrm{1}\right)} \\ $$

Commented by mr W last updated on 29/Dec/21