Solving $2y=sqrt{3+frac{1}{2y}}$
up vote
-1
down vote
favorite
Any way to solve this irrational equation in $mathbb{R}$?
I think it has some artifice, but I do not see it
$$2y=sqrt{3+frac{1}{2y}}$$
algebra-precalculus problem-solving
edited Nov 16 at 11:38
Blue
46.7k870147
asked Nov 16 at 1:42
juan muñoz
84
add a comment |
up vote
-1
down vote
favorite
Any way to solve this irrational equation in $mathbb{R}$?
I think it has some artifice, but I do not see it
$$2y=sqrt{3+frac{1}{2y}}$$
algebra-precalculus problem-solving
edited Nov 16 at 11:38
Blue
46.7k870147
asked Nov 16 at 1:42
juan muñoz
84
Perhaps this question is better placed in stack overflow
– Arvin Singh
Nov 16 at 1:43
Do you mean R or $mathbb R$?
– eyeballfrog
Nov 16 at 1:44
@eyeballfrog R
– juan muñoz
Nov 16 at 2:09
@juanmuñoz: eyeballfrog means R the programming language or $Bbb R$ the real numbers.
– Ross Millikan
Nov 16 at 2:46
@ Ross Millikan .... R the real numbers
– juan muñoz
Nov 17 at 0:26
add a comment |
up vote
-1
down vote
favorite
up vote
-1
down vote
favorite
Any way to solve this irrational equation in $mathbb{R}$?
I think it has some artifice, but I do not see it
$$2y=sqrt{3+frac{1}{2y}}$$
algebra-precalculus problem-solving
edited Nov 16 at 11:38
Blue
46.7k870147
asked Nov 16 at 1:42
juan muñoz
84
Any way to solve this irrational equation in $mathbb{R}$?
I think it has some artifice, but I do not see it
$$2y=sqrt{3+frac{1}{2y}}$$
algebra-precalculus problem-solving
algebra-precalculus problem-solving
edited Nov 16 at 11:38
Blue
46.7k870147
asked Nov 16 at 1:42
juan muñoz
84
edited Nov 16 at 11:38
Blue
46.7k870147
asked Nov 16 at 1:42
juan muñoz
84
edited Nov 16 at 11:38
Blue
46.7k870147
edited Nov 16 at 11:38
Blue
46.7k870147
edited Nov 16 at 11:38
Blue
46.7k870147
46.7k870147
asked Nov 16 at 1:42
juan muñoz
84
asked Nov 16 at 1:42
juan muñoz
84
asked Nov 16 at 1:42
juan muñoz
84
84
Perhaps this question is better placed in stack overflow
– Arvin Singh
Nov 16 at 1:43
Do you mean R or $mathbb R$?
– eyeballfrog
Nov 16 at 1:44
@eyeballfrog R
– juan muñoz
Nov 16 at 2:09
@juanmuñoz: eyeballfrog means R the programming language or $Bbb R$ the real numbers.
– Ross Millikan
Nov 16 at 2:46
@ Ross Millikan .... R the real numbers
– juan muñoz
Nov 17 at 0:26
add a comment |
Perhaps this question is better placed in stack overflow
– Arvin Singh
Nov 16 at 1:43
Do you mean R or $mathbb R$?
– eyeballfrog
Nov 16 at 1:44
@eyeballfrog R
– juan muñoz
Nov 16 at 2:09
@juanmuñoz: eyeballfrog means R the programming language or $Bbb R$ the real numbers.
– Ross Millikan
Nov 16 at 2:46
@ Ross Millikan .... R the real numbers
– juan muñoz
Nov 17 at 0:26
Perhaps this question is better placed in stack overflow
– Arvin Singh
Nov 16 at 1:43
Perhaps this question is better placed in stack overflow
– Arvin Singh
Nov 16 at 1:43
Do you mean R or $mathbb R$?
– eyeballfrog
Nov 16 at 1:44
Do you mean R or $mathbb R$?
– eyeballfrog
Nov 16 at 1:44
@eyeballfrog R
– juan muñoz
Nov 16 at 2:09
@eyeballfrog R
– juan muñoz
Nov 16 at 2:09
@juanmuñoz: eyeballfrog means R the programming language or $Bbb R$ the real numbers.
– Ross Millikan
Nov 16 at 2:46
@juanmuñoz: eyeballfrog means R the programming language or $Bbb R$ the real numbers.
– Ross Millikan
Nov 16 at 2:46
@ Ross Millikan .... R the real numbers
– juan muñoz
Nov 17 at 0:26
@ Ross Millikan .... R the real numbers
– juan muñoz
Nov 17 at 0:26
add a comment |
2 Answers
2
active
oldest
votes
up vote
0
down vote
Square it, clear the fraction, and you have a cubic.
$$8y^3=6y+1$$
Alpha finds three real roots with complicated expressions, about $-0.76604, -0.17365,0.93969$.
answered Nov 16 at 1:51
Ross Millikan
287k23195364
,come to this cubic equation, but how do I solve it by hand
– juan muñoz
Nov 16 at 2:11
add a comment |
up vote
0
down vote
Since $y$ equals to a square root, $y$ is non-negative.
Notice $$2y = sqrt{3+frac{1}{2y}} implies 4y^2 = 3 + frac{1}{2y}
implies 4y^3 - 3y = frac12
$$
and for any $x ge 1$, $4x^3 - 3x ge (4x^2 - 3)x ge 1$, we find $y in (0,1)$.
Take a $theta in (0,frac{pi}{2})$ such that $y = costheta$ and recall $cos(3theta) = 4cos^3theta - 3costheta$. Last equation becomes
$$cos(3theta) = cosfrac{pi}{3}quadimpliesquad 3theta = (2N pm frac13)piquadtext{ for some integer N} $$
Since $theta in (0,frac{pi}{2})$, this forces $theta = frac{pi}{9}$ and hence $y = cosfrac{pi}{9}$
answered Nov 16 at 2:17
achille hui
93.9k5128252
thanks , some way to express cos 20 º with radicals?
– juan muñoz
Nov 17 at 0:25
Yes, but it will involve taking cubic root of complex number which is pretty useless.
– achille hui
Nov 17 at 0:27
You could show me the procedure to learn, if your time allows it
– juan muñoz
Nov 17 at 0:29
That will be the standard Cardano method of solving cubic equation. see entry on wiki and brillant.org
– achille hui
Nov 17 at 0:52
add a comment |
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
0
down vote
Square it, clear the fraction, and you have a cubic.
$$8y^3=6y+1$$
Alpha finds three real roots with complicated expressions, about $-0.76604, -0.17365,0.93969$.
answered Nov 16 at 1:51
Ross Millikan
287k23195364
,come to this cubic equation, but how do I solve it by hand
– juan muñoz
Nov 16 at 2:11
add a comment |
up vote
0
down vote
Square it, clear the fraction, and you have a cubic.
$$8y^3=6y+1$$
Alpha finds three real roots with complicated expressions, about $-0.76604, -0.17365,0.93969$.
answered Nov 16 at 1:51
Ross Millikan
287k23195364
,come to this cubic equation, but how do I solve it by hand
– juan muñoz
Nov 16 at 2:11
add a comment |
up vote
0
down vote
up vote
0
down vote
Square it, clear the fraction, and you have a cubic.
$$8y^3=6y+1$$
Alpha finds three real roots with complicated expressions, about $-0.76604, -0.17365,0.93969$.
answered Nov 16 at 1:51
Ross Millikan
287k23195364
Square it, clear the fraction, and you have a cubic.
$$8y^3=6y+1$$
Alpha finds three real roots with complicated expressions, about $-0.76604, -0.17365,0.93969$.
answered Nov 16 at 1:51
Ross Millikan
287k23195364
answered Nov 16 at 1:51
Ross Millikan
287k23195364
answered Nov 16 at 1:51
Ross Millikan
287k23195364
answered Nov 16 at 1:51
Ross Millikan
287k23195364
287k23195364
,come to this cubic equation, but how do I solve it by hand
– juan muñoz
Nov 16 at 2:11
add a comment |
,come to this cubic equation, but how do I solve it by hand
– juan muñoz
Nov 16 at 2:11
,come to this cubic equation, but how do I solve it by hand
– juan muñoz
Nov 16 at 2:11
,come to this cubic equation, but how do I solve it by hand
– juan muñoz
Nov 16 at 2:11
add a comment |
up vote
0
down vote
Since $y$ equals to a square root, $y$ is non-negative.
Notice $$2y = sqrt{3+frac{1}{2y}} implies 4y^2 = 3 + frac{1}{2y}
implies 4y^3 - 3y = frac12
$$
and for any $x ge 1$, $4x^3 - 3x ge (4x^2 - 3)x ge 1$, we find $y in (0,1)$.
Take a $theta in (0,frac{pi}{2})$ such that $y = costheta$ and recall $cos(3theta) = 4cos^3theta - 3costheta$. Last equation becomes
$$cos(3theta) = cosfrac{pi}{3}quadimpliesquad 3theta = (2N pm frac13)piquadtext{ for some integer N} $$
Since $theta in (0,frac{pi}{2})$, this forces $theta = frac{pi}{9}$ and hence $y = cosfrac{pi}{9}$
answered Nov 16 at 2:17
achille hui
93.9k5128252
thanks , some way to express cos 20 º with radicals?
– juan muñoz
Nov 17 at 0:25
Yes, but it will involve taking cubic root of complex number which is pretty useless.
– achille hui
Nov 17 at 0:27
You could show me the procedure to learn, if your time allows it
– juan muñoz
Nov 17 at 0:29
That will be the standard Cardano method of solving cubic equation. see entry on wiki and brillant.org
– achille hui
Nov 17 at 0:52
add a comment |
up vote
0
down vote
Since $y$ equals to a square root, $y$ is non-negative.
Notice $$2y = sqrt{3+frac{1}{2y}} implies 4y^2 = 3 + frac{1}{2y}
implies 4y^3 - 3y = frac12
$$
and for any $x ge 1$, $4x^3 - 3x ge (4x^2 - 3)x ge 1$, we find $y in (0,1)$.
Take a $theta in (0,frac{pi}{2})$ such that $y = costheta$ and recall $cos(3theta) = 4cos^3theta - 3costheta$. Last equation becomes
$$cos(3theta) = cosfrac{pi}{3}quadimpliesquad 3theta = (2N pm frac13)piquadtext{ for some integer N} $$
Since $theta in (0,frac{pi}{2})$, this forces $theta = frac{pi}{9}$ and hence $y = cosfrac{pi}{9}$
answered Nov 16 at 2:17
achille hui
93.9k5128252
thanks , some way to express cos 20 º with radicals?
– juan muñoz
Nov 17 at 0:25
Yes, but it will involve taking cubic root of complex number which is pretty useless.
– achille hui
Nov 17 at 0:27
You could show me the procedure to learn, if your time allows it
– juan muñoz
Nov 17 at 0:29
That will be the standard Cardano method of solving cubic equation. see entry on wiki and brillant.org
– achille hui
Nov 17 at 0:52
add a comment |
up vote
0
down vote
up vote
0
down vote
Since $y$ equals to a square root, $y$ is non-negative.
Notice $$2y = sqrt{3+frac{1}{2y}} implies 4y^2 = 3 + frac{1}{2y}
implies 4y^3 - 3y = frac12
$$
and for any $x ge 1$, $4x^3 - 3x ge (4x^2 - 3)x ge 1$, we find $y in (0,1)$.
Take a $theta in (0,frac{pi}{2})$ such that $y = costheta$ and recall $cos(3theta) = 4cos^3theta - 3costheta$. Last equation becomes
$$cos(3theta) = cosfrac{pi}{3}quadimpliesquad 3theta = (2N pm frac13)piquadtext{ for some integer N} $$
Since $theta in (0,frac{pi}{2})$, this forces $theta = frac{pi}{9}$ and hence $y = cosfrac{pi}{9}$
answered Nov 16 at 2:17
achille hui
93.9k5128252
Since $y$ equals to a square root, $y$ is non-negative.
Notice $$2y = sqrt{3+frac{1}{2y}} implies 4y^2 = 3 + frac{1}{2y}
implies 4y^3 - 3y = frac12
$$
and for any $x ge 1$, $4x^3 - 3x ge (4x^2 - 3)x ge 1$, we find $y in (0,1)$.
Take a $theta in (0,frac{pi}{2})$ such that $y = costheta$ and recall $cos(3theta) = 4cos^3theta - 3costheta$. Last equation becomes
$$cos(3theta) = cosfrac{pi}{3}quadimpliesquad 3theta = (2N pm frac13)piquadtext{ for some integer N} $$
Since $theta in (0,frac{pi}{2})$, this forces $theta = frac{pi}{9}$ and hence $y = cosfrac{pi}{9}$
answered Nov 16 at 2:17
achille hui
93.9k5128252
answered Nov 16 at 2:17
achille hui
93.9k5128252
answered Nov 16 at 2:17
achille hui
93.9k5128252
answered Nov 16 at 2:17
achille hui
93.9k5128252
93.9k5128252
thanks , some way to express cos 20 º with radicals?
– juan muñoz
Nov 17 at 0:25
Yes, but it will involve taking cubic root of complex number which is pretty useless.
– achille hui
Nov 17 at 0:27
You could show me the procedure to learn, if your time allows it
– juan muñoz
Nov 17 at 0:29
That will be the standard Cardano method of solving cubic equation. see entry on wiki and brillant.org
– achille hui
Nov 17 at 0:52
add a comment |
thanks , some way to express cos 20 º with radicals?
– juan muñoz
Nov 17 at 0:25
Yes, but it will involve taking cubic root of complex number which is pretty useless.
– achille hui
Nov 17 at 0:27
You could show me the procedure to learn, if your time allows it
– juan muñoz
Nov 17 at 0:29
That will be the standard Cardano method of solving cubic equation. see entry on wiki and brillant.org
– achille hui
Nov 17 at 0:52
thanks , some way to express cos 20 º with radicals?
– juan muñoz
Nov 17 at 0:25
thanks , some way to express cos 20 º with radicals?
– juan muñoz
Nov 17 at 0:25
Yes, but it will involve taking cubic root of complex number which is pretty useless.
– achille hui
Nov 17 at 0:27
Yes, but it will involve taking cubic root of complex number which is pretty useless.
– achille hui
Nov 17 at 0:27
You could show me the procedure to learn, if your time allows it
– juan muñoz
Nov 17 at 0:29
You could show me the procedure to learn, if your time allows it
– juan muñoz
Nov 17 at 0:29
That will be the standard Cardano method of solving cubic equation. see entry on wiki and brillant.org
– achille hui
Nov 17 at 0:52
That will be the standard Cardano method of solving cubic equation. see entry on wiki and brillant.org
– achille hui
Nov 17 at 0:52
add a comment |
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3000601%2fsolving-2y-sqrt3-frac12y%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Perhaps this question is better placed in stack overflow
– Arvin Singh
Nov 16 at 1:43
Do you mean R or $mathbb R$?
– eyeballfrog
Nov 16 at 1:44
@eyeballfrog R
– juan muñoz
Nov 16 at 2:09
@juanmuñoz: eyeballfrog means R the programming language or $Bbb R$ the real numbers.
– Ross Millikan
Nov 16 at 2:46
@ Ross Millikan .... R the real numbers
– juan muñoz
Nov 17 at 0:26