How to write $frac{x}{x^2+1}$ in terms of $frac{sqrt{x}}{x+1}$?












0












$begingroup$


I want to rewrite $$frac{x}{x^2+1}$$ in terms of $$frac{sqrt{x}}{x+1}$$
Soo ... I must say
$$frac{sqrt{x}}{x+1}=y$$
Then write $x$ in terms of $y$ ,but how?






algebra-precalculus







share|cite|improve this question









$endgroup$












  • $begingroup$
    Write $x=u^2$ and solve the quadratic equation? [Hint: the final result is $frac{y^2}{1-2 y^2}$]
    $endgroup$
    – AccidentalFourierTransform
    Dec 9 '18 at 19:30


















0












$begingroup$


I want to rewrite $$frac{x}{x^2+1}$$ in terms of $$frac{sqrt{x}}{x+1}$$
Soo ... I must say
$$frac{sqrt{x}}{x+1}=y$$
Then write $x$ in terms of $y$ ,but how?






algebra-precalculus







share|cite|improve this question









$endgroup$












  • $begingroup$
    Write $x=u^2$ and solve the quadratic equation? [Hint: the final result is $frac{y^2}{1-2 y^2}$]
    $endgroup$
    – AccidentalFourierTransform
    Dec 9 '18 at 19:30
















0












0








0





$begingroup$


I want to rewrite $$frac{x}{x^2+1}$$ in terms of $$frac{sqrt{x}}{x+1}$$
Soo ... I must say
$$frac{sqrt{x}}{x+1}=y$$
Then write $x$ in terms of $y$ ,but how?






algebra-precalculus







share|cite|improve this question









$endgroup$




I want to rewrite $$frac{x}{x^2+1}$$ in terms of $$frac{sqrt{x}}{x+1}$$
Soo ... I must say
$$frac{sqrt{x}}{x+1}=y$$
Then write $x$ in terms of $y$ ,but how?






algebra-precalculus




algebra-precalculus






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked Dec 9 '18 at 19:24









user602338user602338

1607




1607












  • $begingroup$
    Write $x=u^2$ and solve the quadratic equation? [Hint: the final result is $frac{y^2}{1-2 y^2}$]
    $endgroup$
    – AccidentalFourierTransform
    Dec 9 '18 at 19:30




















  • $begingroup$
    Write $x=u^2$ and solve the quadratic equation? [Hint: the final result is $frac{y^2}{1-2 y^2}$]
    $endgroup$
    – AccidentalFourierTransform
    Dec 9 '18 at 19:30


















$begingroup$
Write $x=u^2$ and solve the quadratic equation? [Hint: the final result is $frac{y^2}{1-2 y^2}$]
$endgroup$
– AccidentalFourierTransform
Dec 9 '18 at 19:30






$begingroup$
Write $x=u^2$ and solve the quadratic equation? [Hint: the final result is $frac{y^2}{1-2 y^2}$]
$endgroup$
– AccidentalFourierTransform
Dec 9 '18 at 19:30












2 Answers
2






active

oldest

votes


















1












$begingroup$

HINT



We have



$$frac{sqrt{x}}{x+1}=y iff yx-sqrt x+y=0 quad sqrt x=frac{1pmsqrt{1-4y^2}}{2y}$$






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Gimusi sorry. But I am completely confused!
    $endgroup$
    – user602338
    Dec 9 '18 at 19:44










  • $begingroup$
    @user602338 We can consider $yx-sqrt x+y=0$ as a qudratic equation and from here obtain $sqrt x$. It seems that for any $0<y<frac12$we have 2 solutions for $sqrt x$ and one for $y=frac12$. From $sqrt x>0$ we can determine $x$ and $x^2$.
    $endgroup$
    – gimusi
    Dec 9 '18 at 19:48










  • $begingroup$
    Would you please define $a,b,c$ in this quadratic equation?
    $endgroup$
    – user602338
    Dec 9 '18 at 19:50










  • $begingroup$
    @user602338 What do you think they could be $ycdot (sqrt x)^2-1cdot (sqrt x)+y=0$?
    $endgroup$
    – gimusi
    Dec 9 '18 at 19:53










  • $begingroup$
    Oh yes. That's it! Thanks! Just one more problem. How did you find range of $y$ and $sqrt{x}$?
    $endgroup$
    – user602338
    Dec 9 '18 at 19:55





















1












$begingroup$

$$frac{sqrt{x}}{x+1} = y$$ implies $$y^{-2} = x + 2 + x^{-1},$$ so that $$x + x^{-1} = y^{-2} - 2.$$ But $$frac{x}{x^2 + 1} = (x + x^{-1})^{-1} = (y^{-2} - 2)^{-1}.$$






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Can you explain first part of your answer please?
    $endgroup$
    – user602338
    Dec 9 '18 at 19:38










  • $begingroup$
    How did you square both sides?
    $endgroup$
    – user602338
    Dec 9 '18 at 19:38










  • $begingroup$
    @user602338 Why don’t you try showing the first step now that you have seen it? What transformations do you need to apply?
    $endgroup$
    – heropup
    Dec 9 '18 at 19:56










  • $begingroup$
    Ok thanks! I solved the puzzle!
    $endgroup$
    – user602338
    Dec 9 '18 at 19:58










  • $begingroup$
    @user602338 persistence and practice is a big part of learning. Don’t give up so easily. As you have now seen, you figured out how to do the first part, so it is not as if you are totally helpless. You simply need to exercise your skills and your mathematical reasoning will become more sophisticated the more you work at it.
    $endgroup$
    – heropup
    Dec 9 '18 at 20:09











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2 Answers
2






active

oldest

votes








2 Answers
2






active

oldest

votes









active

oldest

votes






active

oldest

votes









1












$begingroup$

HINT



We have



$$frac{sqrt{x}}{x+1}=y iff yx-sqrt x+y=0 quad sqrt x=frac{1pmsqrt{1-4y^2}}{2y}$$






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Gimusi sorry. But I am completely confused!
    $endgroup$
    – user602338
    Dec 9 '18 at 19:44










  • $begingroup$
    @user602338 We can consider $yx-sqrt x+y=0$ as a qudratic equation and from here obtain $sqrt x$. It seems that for any $0<y<frac12$we have 2 solutions for $sqrt x$ and one for $y=frac12$. From $sqrt x>0$ we can determine $x$ and $x^2$.
    $endgroup$
    – gimusi
    Dec 9 '18 at 19:48










  • $begingroup$
    Would you please define $a,b,c$ in this quadratic equation?
    $endgroup$
    – user602338
    Dec 9 '18 at 19:50










  • $begingroup$
    @user602338 What do you think they could be $ycdot (sqrt x)^2-1cdot (sqrt x)+y=0$?
    $endgroup$
    – gimusi
    Dec 9 '18 at 19:53










  • $begingroup$
    Oh yes. That's it! Thanks! Just one more problem. How did you find range of $y$ and $sqrt{x}$?
    $endgroup$
    – user602338
    Dec 9 '18 at 19:55


















1












$begingroup$

HINT



We have



$$frac{sqrt{x}}{x+1}=y iff yx-sqrt x+y=0 quad sqrt x=frac{1pmsqrt{1-4y^2}}{2y}$$






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Gimusi sorry. But I am completely confused!
    $endgroup$
    – user602338
    Dec 9 '18 at 19:44










  • $begingroup$
    @user602338 We can consider $yx-sqrt x+y=0$ as a qudratic equation and from here obtain $sqrt x$. It seems that for any $0<y<frac12$we have 2 solutions for $sqrt x$ and one for $y=frac12$. From $sqrt x>0$ we can determine $x$ and $x^2$.
    $endgroup$
    – gimusi
    Dec 9 '18 at 19:48










  • $begingroup$
    Would you please define $a,b,c$ in this quadratic equation?
    $endgroup$
    – user602338
    Dec 9 '18 at 19:50










  • $begingroup$
    @user602338 What do you think they could be $ycdot (sqrt x)^2-1cdot (sqrt x)+y=0$?
    $endgroup$
    – gimusi
    Dec 9 '18 at 19:53










  • $begingroup$
    Oh yes. That's it! Thanks! Just one more problem. How did you find range of $y$ and $sqrt{x}$?
    $endgroup$
    – user602338
    Dec 9 '18 at 19:55
















1












1








1





$begingroup$

HINT



We have



$$frac{sqrt{x}}{x+1}=y iff yx-sqrt x+y=0 quad sqrt x=frac{1pmsqrt{1-4y^2}}{2y}$$






share|cite|improve this answer









$endgroup$



HINT



We have



$$frac{sqrt{x}}{x+1}=y iff yx-sqrt x+y=0 quad sqrt x=frac{1pmsqrt{1-4y^2}}{2y}$$







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Dec 9 '18 at 19:37









gimusigimusi

92.8k84494




92.8k84494












  • $begingroup$
    Gimusi sorry. But I am completely confused!
    $endgroup$
    – user602338
    Dec 9 '18 at 19:44










  • $begingroup$
    @user602338 We can consider $yx-sqrt x+y=0$ as a qudratic equation and from here obtain $sqrt x$. It seems that for any $0<y<frac12$we have 2 solutions for $sqrt x$ and one for $y=frac12$. From $sqrt x>0$ we can determine $x$ and $x^2$.
    $endgroup$
    – gimusi
    Dec 9 '18 at 19:48










  • $begingroup$
    Would you please define $a,b,c$ in this quadratic equation?
    $endgroup$
    – user602338
    Dec 9 '18 at 19:50










  • $begingroup$
    @user602338 What do you think they could be $ycdot (sqrt x)^2-1cdot (sqrt x)+y=0$?
    $endgroup$
    – gimusi
    Dec 9 '18 at 19:53










  • $begingroup$
    Oh yes. That's it! Thanks! Just one more problem. How did you find range of $y$ and $sqrt{x}$?
    $endgroup$
    – user602338
    Dec 9 '18 at 19:55




















  • $begingroup$
    Gimusi sorry. But I am completely confused!
    $endgroup$
    – user602338
    Dec 9 '18 at 19:44










  • $begingroup$
    @user602338 We can consider $yx-sqrt x+y=0$ as a qudratic equation and from here obtain $sqrt x$. It seems that for any $0<y<frac12$we have 2 solutions for $sqrt x$ and one for $y=frac12$. From $sqrt x>0$ we can determine $x$ and $x^2$.
    $endgroup$
    – gimusi
    Dec 9 '18 at 19:48










  • $begingroup$
    Would you please define $a,b,c$ in this quadratic equation?
    $endgroup$
    – user602338
    Dec 9 '18 at 19:50










  • $begingroup$
    @user602338 What do you think they could be $ycdot (sqrt x)^2-1cdot (sqrt x)+y=0$?
    $endgroup$
    – gimusi
    Dec 9 '18 at 19:53










  • $begingroup$
    Oh yes. That's it! Thanks! Just one more problem. How did you find range of $y$ and $sqrt{x}$?
    $endgroup$
    – user602338
    Dec 9 '18 at 19:55


















$begingroup$
Gimusi sorry. But I am completely confused!
$endgroup$
– user602338
Dec 9 '18 at 19:44




$begingroup$
Gimusi sorry. But I am completely confused!
$endgroup$
– user602338
Dec 9 '18 at 19:44












$begingroup$
@user602338 We can consider $yx-sqrt x+y=0$ as a qudratic equation and from here obtain $sqrt x$. It seems that for any $0<y<frac12$we have 2 solutions for $sqrt x$ and one for $y=frac12$. From $sqrt x>0$ we can determine $x$ and $x^2$.
$endgroup$
– gimusi
Dec 9 '18 at 19:48




$begingroup$
@user602338 We can consider $yx-sqrt x+y=0$ as a qudratic equation and from here obtain $sqrt x$. It seems that for any $0<y<frac12$we have 2 solutions for $sqrt x$ and one for $y=frac12$. From $sqrt x>0$ we can determine $x$ and $x^2$.
$endgroup$
– gimusi
Dec 9 '18 at 19:48












$begingroup$
Would you please define $a,b,c$ in this quadratic equation?
$endgroup$
– user602338
Dec 9 '18 at 19:50




$begingroup$
Would you please define $a,b,c$ in this quadratic equation?
$endgroup$
– user602338
Dec 9 '18 at 19:50












$begingroup$
@user602338 What do you think they could be $ycdot (sqrt x)^2-1cdot (sqrt x)+y=0$?
$endgroup$
– gimusi
Dec 9 '18 at 19:53




$begingroup$
@user602338 What do you think they could be $ycdot (sqrt x)^2-1cdot (sqrt x)+y=0$?
$endgroup$
– gimusi
Dec 9 '18 at 19:53












$begingroup$
Oh yes. That's it! Thanks! Just one more problem. How did you find range of $y$ and $sqrt{x}$?
$endgroup$
– user602338
Dec 9 '18 at 19:55






$begingroup$
Oh yes. That's it! Thanks! Just one more problem. How did you find range of $y$ and $sqrt{x}$?
$endgroup$
– user602338
Dec 9 '18 at 19:55













1












$begingroup$

$$frac{sqrt{x}}{x+1} = y$$ implies $$y^{-2} = x + 2 + x^{-1},$$ so that $$x + x^{-1} = y^{-2} - 2.$$ But $$frac{x}{x^2 + 1} = (x + x^{-1})^{-1} = (y^{-2} - 2)^{-1}.$$






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Can you explain first part of your answer please?
    $endgroup$
    – user602338
    Dec 9 '18 at 19:38










  • $begingroup$
    How did you square both sides?
    $endgroup$
    – user602338
    Dec 9 '18 at 19:38










  • $begingroup$
    @user602338 Why don’t you try showing the first step now that you have seen it? What transformations do you need to apply?
    $endgroup$
    – heropup
    Dec 9 '18 at 19:56










  • $begingroup$
    Ok thanks! I solved the puzzle!
    $endgroup$
    – user602338
    Dec 9 '18 at 19:58










  • $begingroup$
    @user602338 persistence and practice is a big part of learning. Don’t give up so easily. As you have now seen, you figured out how to do the first part, so it is not as if you are totally helpless. You simply need to exercise your skills and your mathematical reasoning will become more sophisticated the more you work at it.
    $endgroup$
    – heropup
    Dec 9 '18 at 20:09
















1












$begingroup$

$$frac{sqrt{x}}{x+1} = y$$ implies $$y^{-2} = x + 2 + x^{-1},$$ so that $$x + x^{-1} = y^{-2} - 2.$$ But $$frac{x}{x^2 + 1} = (x + x^{-1})^{-1} = (y^{-2} - 2)^{-1}.$$






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Can you explain first part of your answer please?
    $endgroup$
    – user602338
    Dec 9 '18 at 19:38










  • $begingroup$
    How did you square both sides?
    $endgroup$
    – user602338
    Dec 9 '18 at 19:38










  • $begingroup$
    @user602338 Why don’t you try showing the first step now that you have seen it? What transformations do you need to apply?
    $endgroup$
    – heropup
    Dec 9 '18 at 19:56










  • $begingroup$
    Ok thanks! I solved the puzzle!
    $endgroup$
    – user602338
    Dec 9 '18 at 19:58










  • $begingroup$
    @user602338 persistence and practice is a big part of learning. Don’t give up so easily. As you have now seen, you figured out how to do the first part, so it is not as if you are totally helpless. You simply need to exercise your skills and your mathematical reasoning will become more sophisticated the more you work at it.
    $endgroup$
    – heropup
    Dec 9 '18 at 20:09














1












1








1





$begingroup$

$$frac{sqrt{x}}{x+1} = y$$ implies $$y^{-2} = x + 2 + x^{-1},$$ so that $$x + x^{-1} = y^{-2} - 2.$$ But $$frac{x}{x^2 + 1} = (x + x^{-1})^{-1} = (y^{-2} - 2)^{-1}.$$






share|cite|improve this answer









$endgroup$



$$frac{sqrt{x}}{x+1} = y$$ implies $$y^{-2} = x + 2 + x^{-1},$$ so that $$x + x^{-1} = y^{-2} - 2.$$ But $$frac{x}{x^2 + 1} = (x + x^{-1})^{-1} = (y^{-2} - 2)^{-1}.$$







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Dec 9 '18 at 19:34









heropupheropup

63.6k762102




63.6k762102












  • $begingroup$
    Can you explain first part of your answer please?
    $endgroup$
    – user602338
    Dec 9 '18 at 19:38










  • $begingroup$
    How did you square both sides?
    $endgroup$
    – user602338
    Dec 9 '18 at 19:38










  • $begingroup$
    @user602338 Why don’t you try showing the first step now that you have seen it? What transformations do you need to apply?
    $endgroup$
    – heropup
    Dec 9 '18 at 19:56










  • $begingroup$
    Ok thanks! I solved the puzzle!
    $endgroup$
    – user602338
    Dec 9 '18 at 19:58










  • $begingroup$
    @user602338 persistence and practice is a big part of learning. Don’t give up so easily. As you have now seen, you figured out how to do the first part, so it is not as if you are totally helpless. You simply need to exercise your skills and your mathematical reasoning will become more sophisticated the more you work at it.
    $endgroup$
    – heropup
    Dec 9 '18 at 20:09


















  • $begingroup$
    Can you explain first part of your answer please?
    $endgroup$
    – user602338
    Dec 9 '18 at 19:38










  • $begingroup$
    How did you square both sides?
    $endgroup$
    – user602338
    Dec 9 '18 at 19:38










  • $begingroup$
    @user602338 Why don’t you try showing the first step now that you have seen it? What transformations do you need to apply?
    $endgroup$
    – heropup
    Dec 9 '18 at 19:56










  • $begingroup$
    Ok thanks! I solved the puzzle!
    $endgroup$
    – user602338
    Dec 9 '18 at 19:58










  • $begingroup$
    @user602338 persistence and practice is a big part of learning. Don’t give up so easily. As you have now seen, you figured out how to do the first part, so it is not as if you are totally helpless. You simply need to exercise your skills and your mathematical reasoning will become more sophisticated the more you work at it.
    $endgroup$
    – heropup
    Dec 9 '18 at 20:09
















$begingroup$
Can you explain first part of your answer please?
$endgroup$
– user602338
Dec 9 '18 at 19:38




$begingroup$
Can you explain first part of your answer please?
$endgroup$
– user602338
Dec 9 '18 at 19:38












$begingroup$
How did you square both sides?
$endgroup$
– user602338
Dec 9 '18 at 19:38




$begingroup$
How did you square both sides?
$endgroup$
– user602338
Dec 9 '18 at 19:38












$begingroup$
@user602338 Why don’t you try showing the first step now that you have seen it? What transformations do you need to apply?
$endgroup$
– heropup
Dec 9 '18 at 19:56




$begingroup$
@user602338 Why don’t you try showing the first step now that you have seen it? What transformations do you need to apply?
$endgroup$
– heropup
Dec 9 '18 at 19:56












$begingroup$
Ok thanks! I solved the puzzle!
$endgroup$
– user602338
Dec 9 '18 at 19:58




$begingroup$
Ok thanks! I solved the puzzle!
$endgroup$
– user602338
Dec 9 '18 at 19:58












$begingroup$
@user602338 persistence and practice is a big part of learning. Don’t give up so easily. As you have now seen, you figured out how to do the first part, so it is not as if you are totally helpless. You simply need to exercise your skills and your mathematical reasoning will become more sophisticated the more you work at it.
$endgroup$
– heropup
Dec 9 '18 at 20:09




$begingroup$
@user602338 persistence and practice is a big part of learning. Don’t give up so easily. As you have now seen, you figured out how to do the first part, so it is not as if you are totally helpless. You simply need to exercise your skills and your mathematical reasoning will become more sophisticated the more you work at it.
$endgroup$
– heropup
Dec 9 '18 at 20:09


















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Mont Emei

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Mont Emei.mw-parser-output .entete.map{background-image:url("//upload.wikimedia.org/wikipedia/commons/7/7a/Picto_infobox_map.png")} Vue du sommet Wanfo du mont Emei Géographie Altitude 3 099  m , Wanfo Coordonnées 29° 31′ 11″ nord, 103° 19′ 57″ est Administration Pays   Chine Province Sichuan Ville-préfecture Leshan Géolocalisation sur la carte : Sichuan Mont Emei Géolocalisation sur la carte : Chine Mont Emei modifier   Paysage panoramique du mont Emei, incluant le paysage panoramique du grand Bouddha de Leshan *   Patrimoine mondial de l'UNESCO Passerelle en bois sur le versant ouest Pays   Chine Subdivision Sichuan Type Mixte Critères (iv) (vi) (x) Superficie 15 400 ha Numéro d’identification 779 Zone géographique Asie et Pacifique  ** Année d’inscription 1996 (20 e session) * Descriptif officiel UNESCO ** Classification géog
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