Analytical solution to $sqrta^2-x^2 + sqrtb^2-x^2 = sqrta^2-x^2 cdot sqrtb^2-x^2$ The 2019 Stack Overflow Developer Survey Results Are In Announcing the arrival of Valued Associate #679: Cesar Manara Planned maintenance scheduled April 17/18, 2019 at 00:00UTC (8:00pm US/Eastern)Analytical method for root findingReal solutions of the equation $x = sqrt3-x cdot sqrt4-x + sqrt4-x cdot sqrt5-x + sqrt5-x cdot sqrt3-x$Is there a general rule for proving that an equation has no analyticial solutionAnalytical solution for rational equality, square root in denominator on both sidesanalytical solution to equation$sqrtx^2-1 =sqrtx+1cdotsqrtx-1.$Lambert function - Need help finding an analytical solutionSolve $lfloor sqrt x +sqrtx+1+sqrtx+2rfloor=x$How to confidently and concretely state that a problem has no analytical solutionAnalytical solution to the crossed ladders problem
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Analytical solution to $sqrta^2-x^2 + sqrtb^2-x^2 = sqrta^2-x^2 cdot sqrtb^2-x^2$
The 2019 Stack Overflow Developer Survey Results Are In
Announcing the arrival of Valued Associate #679: Cesar Manara
Planned maintenance scheduled April 17/18, 2019 at 00:00UTC (8:00pm US/Eastern)Analytical method for root findingReal solutions of the equation $x = sqrt3-x cdot sqrt4-x + sqrt4-x cdot sqrt5-x + sqrt5-x cdot sqrt3-x$Is there a general rule for proving that an equation has no analyticial solutionAnalytical solution for rational equality, square root in denominator on both sidesanalytical solution to equation$sqrtx^2-1 =sqrtx+1cdotsqrtx-1.$Lambert function - Need help finding an analytical solutionSolve $lfloor sqrt x +sqrtx+1+sqrtx+2rfloor=x$How to confidently and concretely state that a problem has no analytical solutionAnalytical solution to the crossed ladders problem
$begingroup$
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
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.
$endgroup$
|
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
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
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.
$endgroup$
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
algebra-precalculus
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
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
edited Apr 8 at 13:52
Lok
404
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
asked Apr 8 at 13:25
UtheringUthering
61
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.
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.
Uthering is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
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
$begingroup$
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
1
1
$begingroup$
What is $N_0$ ?
$endgroup$
– Rebellos
Apr 8 at 13:33
$begingroup$
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
$begingroup$
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
2
$begingroup$
This is similar to solving $x+y=xy$.
$endgroup$
– Vasya
Apr 8 at 13:37
$begingroup$
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
$begingroup$
@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
$begingroup$
$N_0$ is naturals with zero, i.e. 0, 1, 2, ...
$endgroup$
– Uthering
Apr 8 at 13:47
|
show 2 more comments
2 Answers
2
active
oldest
votes
$begingroup$
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.
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 |
$begingroup$
So, we have $$dfrac1sqrta^2-x^2+dfrac1sqrtb^2-x^2=1$$
WLOG $a^2-x^2=sin^4t, b^2-x^2=cos^4t$
$implies b^2-a^2=cos2t$
$$4a^2-4x^2=(2sin^2t)^2=(1+a^2-b^2)^2$$
$$iff4x^2=?$$
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 |
Your Answer
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
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active
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active
oldest
votes
$begingroup$
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.
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.
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.
answered Apr 8 at 13:43
Yves DaoustYves Daoust
133k676231
$endgroup$
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.
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
answered Apr 8 at 13:43
Yves DaoustYves Daoust
133k676231
133k676231
$begingroup$
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 |
$begingroup$
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
$begingroup$
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
$begingroup$
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
$begingroup$
@Uthering: it does.
$endgroup$
– Yves Daoust
Apr 8 at 15:15
add a comment |
$begingroup$
So, we have $$dfrac1sqrta^2-x^2+dfrac1sqrtb^2-x^2=1$$
WLOG $a^2-x^2=sin^4t, b^2-x^2=cos^4t$
$implies b^2-a^2=cos2t$
$$4a^2-4x^2=(2sin^2t)^2=(1+a^2-b^2)^2$$
$$iff4x^2=?$$
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 $a^2-x^2=sin^4t, b^2-x^2=cos^4t$
$implies b^2-a^2=cos2t$
$$4a^2-4x^2=(2sin^2t)^2=(1+a^2-b^2)^2$$
$$iff4x^2=?$$
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 $a^2-x^2=sin^4t, b^2-x^2=cos^4t$
$implies b^2-a^2=cos2t$
$$4a^2-4x^2=(2sin^2t)^2=(1+a^2-b^2)^2$$
$$iff4x^2=?$$
answered Apr 8 at 13:58
lab bhattacharjeelab bhattacharjee
228k15159279
$endgroup$
So, we have $$dfrac1sqrta^2-x^2+dfrac1sqrtb^2-x^2=1$$
WLOG $a^2-x^2=sin^4t, b^2-x^2=cos^4t$
$implies b^2-a^2=cos2t$
$$4a^2-4x^2=(2sin^2t)^2=(1+a^2-b^2)^2$$
$$iff4x^2=?$$
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
answered Apr 8 at 13:58
lab bhattacharjeelab bhattacharjee
228k15159279
228k15159279
$begingroup$
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 |
$begingroup$
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
$begingroup$
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
$begingroup$
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 |
Uthering is a new contributor. Be nice, and check out our Code of Conduct.
Uthering is a new contributor. Be nice, and check out our Code of Conduct.
Uthering is a new contributor. Be nice, and check out our Code of Conduct.
Uthering is a new contributor. Be nice, and check out our Code of Conduct.
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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