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Analytical solution to $sqrta^2-x^2 + sqrtb^2-x^2 = sqrta^2-x^2 cdot sqrtb^2-x^2$

1

2

$begingroup$

Can't find beauty analytical solution to such equation :

$$sqrta^2-x^2 + sqrtb^2-x^2 = sqrta^2-x^2cdotsqrtb^2-x^2$$

Assuming $a, b in mathbbN, a le b, x in mathbbR$

Is it possible to find a solution for general case?

algebra-precalculus

share|cite|improve this question

edited Apr 8 at 13:52

Lok

404

asked Apr 8 at 13:25

UtheringUthering

61

New contributor

Uthering is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

$endgroup$

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  What is $N_0$ ?
  $endgroup$
  – Rebellos
  Apr 8 at 13:33
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Welcome to stackexchange. Please edit the question to show us what you tried and where you are stuck.
  $endgroup$
  – Ethan Bolker
  Apr 8 at 13:36
  

  
  
  
  


• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  $begingroup$
  This is similar to solving $x+y=xy$.
  $endgroup$
  – Vasya
  Apr 8 at 13:37
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @Vasya: how would you solve $x+y=xy$ ??
  $endgroup$
  – Yves Daoust
  Apr 8 at 13:44
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  $N_0$ is naturals with zero, i.e. 0, 1, 2, ...
  $endgroup$
  – Uthering
  Apr 8 at 13:47
  

  
  
  
  

 | 
show 2 more comments

1

2

$begingroup$

Can't find beauty analytical solution to such equation :

$$sqrta^2-x^2 + sqrtb^2-x^2 = sqrta^2-x^2cdotsqrtb^2-x^2$$

Assuming $a, b in mathbbN, a le b, x in mathbbR$

Is it possible to find a solution for general case?

algebra-precalculus

share|cite|improve this question

edited Apr 8 at 13:52

Lok

404

asked Apr 8 at 13:25

UtheringUthering

61

New contributor

Uthering is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

$endgroup$

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  What is $N_0$ ?
  $endgroup$
  – Rebellos
  Apr 8 at 13:33
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Welcome to stackexchange. Please edit the question to show us what you tried and where you are stuck.
  $endgroup$
  – Ethan Bolker
  Apr 8 at 13:36
  

  
  
  
  


• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  $begingroup$
  This is similar to solving $x+y=xy$.
  $endgroup$
  – Vasya
  Apr 8 at 13:37
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @Vasya: how would you solve $x+y=xy$ ??
  $endgroup$
  – Yves Daoust
  Apr 8 at 13:44
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  $N_0$ is naturals with zero, i.e. 0, 1, 2, ...
  $endgroup$
  – Uthering
  Apr 8 at 13:47
  

  
  
  
  

 | 
show 2 more comments

$begingroup$

Can't find beauty analytical solution to such equation :

$$sqrta^2-x^2 + sqrtb^2-x^2 = sqrta^2-x^2cdotsqrtb^2-x^2$$

Assuming $a, b in mathbbN, a le b, x in mathbbR$

Is it possible to find a solution for general case?

algebra-precalculus

share|cite|improve this question

edited Apr 8 at 13:52

Lok

404

asked Apr 8 at 13:25

UtheringUthering

61

New contributor

Uthering is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

$endgroup$

share|cite|improve this question

edited Apr 8 at 13:52

Lok

404

asked Apr 8 at 13:25

UtheringUthering

61

New contributor

Uthering is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

share|cite|improve this question

edited Apr 8 at 13:52

Lok

404

asked Apr 8 at 13:25

UtheringUthering

61

New contributor

Uthering is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

edited Apr 8 at 13:52

Lok

404

edited Apr 8 at 13:52

Lok

404

asked Apr 8 at 13:25

UtheringUthering

61

New contributor

Uthering is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

asked Apr 8 at 13:25

UtheringUthering

61

2 Answers
2

active

oldest

votes

0

$begingroup$

Hint:

Let $sqrta^2-x^2=y$. We have

$$y+sqrtb^2-a^2+y^2=ysqrtb^2-a^2+y^2,$$

$$y=(y-1)sqrtb^2-a^2+y^2.$$

Squaring, you will obtain a quartic equation, which doesn't seem to simplify.

share|cite|improve this answer

answered Apr 8 at 13:43

Yves DaoustYves Daoust

133k676231

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Yes, a good hint, not tried it yet. I've tried to solve quartic -- it looks very ugly, yet :/
  $endgroup$
  – Uthering
  Apr 8 at 13:48
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @Uthering: it does.
  $endgroup$
  – Yves Daoust
  Apr 8 at 15:15
  

  
  
  
  

add a comment | 

0

$begingroup$

So, we have $$dfrac1sqrta^2-x^2+dfrac1sqrtb^2-x^2=1$$

WLOG $a^2-x^2=sin^4t, b^2-x^2=cos^4t$

$implies b^2-a^2=cos2t$

$$4a^2-4x^2=(2sin^2t)^2=(1+a^2-b^2)^2$$

$$iff4x^2=?$$

share|cite|improve this answer

answered Apr 8 at 13:58

lab bhattacharjeelab bhattacharjee

228k15159279

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Looks very nice :) Yet, $a^2-x^2=sin^4t$ states that $a^2-x^2 in [-1,1]$ and we loose solutions. If I state that, f.e. $a^2-x^2=r^2*sin^4t$, than again terrible thing appears :/
  $endgroup$
  – Uthering
  Apr 9 at 6:26
  
  

  
  
  
  

add a comment | 

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0

$begingroup$

Hint:

Let $sqrta^2-x^2=y$. We have

$$y+sqrtb^2-a^2+y^2=ysqrtb^2-a^2+y^2,$$

$$y=(y-1)sqrtb^2-a^2+y^2.$$

Squaring, you will obtain a quartic equation, which doesn't seem to simplify.

share|cite|improve this answer

answered Apr 8 at 13:43

Yves DaoustYves Daoust

133k676231

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Yes, a good hint, not tried it yet. I've tried to solve quartic -- it looks very ugly, yet :/
  $endgroup$
  – Uthering
  Apr 8 at 13:48
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @Uthering: it does.
  $endgroup$
  – Yves Daoust
  Apr 8 at 15:15
  

  
  
  
  

add a comment | 

0

$begingroup$

Hint:

Let $sqrta^2-x^2=y$. We have

$$y+sqrtb^2-a^2+y^2=ysqrtb^2-a^2+y^2,$$

$$y=(y-1)sqrtb^2-a^2+y^2.$$

Squaring, you will obtain a quartic equation, which doesn't seem to simplify.

share|cite|improve this answer

answered Apr 8 at 13:43

Yves DaoustYves Daoust

133k676231

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Yes, a good hint, not tried it yet. I've tried to solve quartic -- it looks very ugly, yet :/
  $endgroup$
  – Uthering
  Apr 8 at 13:48
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @Uthering: it does.
  $endgroup$
  – Yves Daoust
  Apr 8 at 15:15
  

  
  
  
  

add a comment | 

$begingroup$

Hint:

Let $sqrta^2-x^2=y$. We have

$$y+sqrtb^2-a^2+y^2=ysqrtb^2-a^2+y^2,$$

$$y=(y-1)sqrtb^2-a^2+y^2.$$

Squaring, you will obtain a quartic equation, which doesn't seem to simplify.

share|cite|improve this answer

answered Apr 8 at 13:43

Yves DaoustYves Daoust

133k676231

$endgroup$

share|cite|improve this answer

answered Apr 8 at 13:43

Yves DaoustYves Daoust

133k676231

answered Apr 8 at 13:43

Yves DaoustYves Daoust

133k676231

answered Apr 8 at 13:43

Yves DaoustYves Daoust

133k676231

0

$begingroup$

So, we have $$dfrac1sqrta^2-x^2+dfrac1sqrtb^2-x^2=1$$

WLOG $a^2-x^2=sin^4t, b^2-x^2=cos^4t$

$implies b^2-a^2=cos2t$

$$4a^2-4x^2=(2sin^2t)^2=(1+a^2-b^2)^2$$

$$iff4x^2=?$$

share|cite|improve this answer

answered Apr 8 at 13:58

lab bhattacharjeelab bhattacharjee

228k15159279

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Looks very nice :) Yet, $a^2-x^2=sin^4t$ states that $a^2-x^2 in [-1,1]$ and we loose solutions. If I state that, f.e. $a^2-x^2=r^2*sin^4t$, than again terrible thing appears :/
  $endgroup$
  – Uthering
  Apr 9 at 6:26
  
  

  
  
  
  

add a comment | 

0

$begingroup$

So, we have $$dfrac1sqrta^2-x^2+dfrac1sqrtb^2-x^2=1$$

WLOG $a^2-x^2=sin^4t, b^2-x^2=cos^4t$

$implies b^2-a^2=cos2t$

$$4a^2-4x^2=(2sin^2t)^2=(1+a^2-b^2)^2$$

$$iff4x^2=?$$

share|cite|improve this answer

answered Apr 8 at 13:58

lab bhattacharjeelab bhattacharjee

228k15159279

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Looks very nice :) Yet, $a^2-x^2=sin^4t$ states that $a^2-x^2 in [-1,1]$ and we loose solutions. If I state that, f.e. $a^2-x^2=r^2*sin^4t$, than again terrible thing appears :/
  $endgroup$
  – Uthering
  Apr 9 at 6:26
  
  

  
  
  
  

add a comment | 

$begingroup$

So, we have $$dfrac1sqrta^2-x^2+dfrac1sqrtb^2-x^2=1$$

WLOG $a^2-x^2=sin^4t, b^2-x^2=cos^4t$

$implies b^2-a^2=cos2t$

$$4a^2-4x^2=(2sin^2t)^2=(1+a^2-b^2)^2$$

$$iff4x^2=?$$

share|cite|improve this answer

answered Apr 8 at 13:58

lab bhattacharjeelab bhattacharjee

228k15159279

$endgroup$

share|cite|improve this answer

answered Apr 8 at 13:58

lab bhattacharjeelab bhattacharjee

228k15159279

answered Apr 8 at 13:58

lab bhattacharjeelab bhattacharjee

228k15159279

answered Apr 8 at 13:58

lab bhattacharjeelab bhattacharjee

228k15159279