Solving $a x^4 + b x + c = 0$
up vote
3
down vote
favorite
This question is about the methods to solve polynomials of the fourth degree with the form $ax^4 + bx + c = 0$.
I am currently creating an algorithm that needs to calculate the solution of the equation $a T^4 + bT + c = 0$, with $T$ being the temperature.
Obviously, temperatures are positive reals, so I need the only positive real root.
In the algorithm of MATLAB, I use a function to solve any polynomial; but I would like to compare it to a direct formula. (Empirically, there is always one and only one positive real root, but I haven't proved it)
So I went to look up for the equation of the solutions on Wolfram Alpha to get the general root expressions and try to compute it directly.
However, I always ended up with complex results, I don't know exactly why.
Also the solution provided by Wolfram Alpha doesn't show where the square root sign ends, so I am pretty sure I messed that up. So what is the general expression for the equation $ax^4 + bx + c = 0$ (an expression in which it is clear where the operators start and end; or using temporary variables to calculate it)?
Finally, I tried to check some methods to solve fourth-degree polynomials and tried to use the Ferrari method, but I didn't end up getting a real root.
Are there methods that don't required "advanced" calculating methods (MATLAB root function), but ones based on elementary operations?
polynomials
edited Nov 17 at 12:39
amWhy
191k27223438
asked Nov 15 at 11:48
PackSciences
42916
add a comment |
up vote
3
down vote
favorite
This question is about the methods to solve polynomials of the fourth degree with the form $ax^4 + bx + c = 0$.
I am currently creating an algorithm that needs to calculate the solution of the equation $a T^4 + bT + c = 0$, with $T$ being the temperature.
Obviously, temperatures are positive reals, so I need the only positive real root.
In the algorithm of MATLAB, I use a function to solve any polynomial; but I would like to compare it to a direct formula. (Empirically, there is always one and only one positive real root, but I haven't proved it)
So I went to look up for the equation of the solutions on Wolfram Alpha to get the general root expressions and try to compute it directly.
However, I always ended up with complex results, I don't know exactly why.
Also the solution provided by Wolfram Alpha doesn't show where the square root sign ends, so I am pretty sure I messed that up. So what is the general expression for the equation $ax^4 + bx + c = 0$ (an expression in which it is clear where the operators start and end; or using temporary variables to calculate it)?
Finally, I tried to check some methods to solve fourth-degree polynomials and tried to use the Ferrari method, but I didn't end up getting a real root.
Are there methods that don't required "advanced" calculating methods (MATLAB root function), but ones based on elementary operations?
polynomials
edited Nov 17 at 12:39
amWhy
191k27223438
asked Nov 15 at 11:48
PackSciences
42916
2
this link may be helpful: math.stackexchange.com/questions/785/…
– Sujit Bhattacharyya
Nov 15 at 11:52
Computing the discriminant and using Descartes' Rule of Signs shows that unless $c = 0$, this equation either has $0$ or $2$ real roots, according to whether $256 a c^3 - 27 b^4$ is positive or negative, respectively. (If the discriminant is zero, either is possible.)
– Travis
Nov 15 at 12:28
2
At any rate, the general formula is somewhat unpleasant, and probably for your (empirical) setting a numerical solution is easier and more reasonable besides.
– Travis
Nov 15 at 12:29
add a comment |
up vote
3
down vote
favorite
up vote
3
down vote
favorite
This question is about the methods to solve polynomials of the fourth degree with the form $ax^4 + bx + c = 0$.
I am currently creating an algorithm that needs to calculate the solution of the equation $a T^4 + bT + c = 0$, with $T$ being the temperature.
Obviously, temperatures are positive reals, so I need the only positive real root.
In the algorithm of MATLAB, I use a function to solve any polynomial; but I would like to compare it to a direct formula. (Empirically, there is always one and only one positive real root, but I haven't proved it)
So I went to look up for the equation of the solutions on Wolfram Alpha to get the general root expressions and try to compute it directly.
However, I always ended up with complex results, I don't know exactly why.
Also the solution provided by Wolfram Alpha doesn't show where the square root sign ends, so I am pretty sure I messed that up. So what is the general expression for the equation $ax^4 + bx + c = 0$ (an expression in which it is clear where the operators start and end; or using temporary variables to calculate it)?
Finally, I tried to check some methods to solve fourth-degree polynomials and tried to use the Ferrari method, but I didn't end up getting a real root.
Are there methods that don't required "advanced" calculating methods (MATLAB root function), but ones based on elementary operations?
polynomials
edited Nov 17 at 12:39
amWhy
191k27223438
asked Nov 15 at 11:48
PackSciences
42916
This question is about the methods to solve polynomials of the fourth degree with the form $ax^4 + bx + c = 0$.
I am currently creating an algorithm that needs to calculate the solution of the equation $a T^4 + bT + c = 0$, with $T$ being the temperature.
Obviously, temperatures are positive reals, so I need the only positive real root.
In the algorithm of MATLAB, I use a function to solve any polynomial; but I would like to compare it to a direct formula. (Empirically, there is always one and only one positive real root, but I haven't proved it)
So I went to look up for the equation of the solutions on Wolfram Alpha to get the general root expressions and try to compute it directly.
However, I always ended up with complex results, I don't know exactly why.
Also the solution provided by Wolfram Alpha doesn't show where the square root sign ends, so I am pretty sure I messed that up. So what is the general expression for the equation $ax^4 + bx + c = 0$ (an expression in which it is clear where the operators start and end; or using temporary variables to calculate it)?
Finally, I tried to check some methods to solve fourth-degree polynomials and tried to use the Ferrari method, but I didn't end up getting a real root.
Are there methods that don't required "advanced" calculating methods (MATLAB root function), but ones based on elementary operations?
polynomials
polynomials
edited Nov 17 at 12:39
amWhy
191k27223438
asked Nov 15 at 11:48
PackSciences
42916
edited Nov 17 at 12:39
amWhy
191k27223438
asked Nov 15 at 11:48
PackSciences
42916
edited Nov 17 at 12:39
amWhy
191k27223438
edited Nov 17 at 12:39
amWhy
191k27223438
edited Nov 17 at 12:39
amWhy
191k27223438
191k27223438
asked Nov 15 at 11:48
PackSciences
42916
asked Nov 15 at 11:48
PackSciences
42916
asked Nov 15 at 11:48
PackSciences
42916
42916
2
this link may be helpful: math.stackexchange.com/questions/785/…
– Sujit Bhattacharyya
Nov 15 at 11:52
Computing the discriminant and using Descartes' Rule of Signs shows that unless $c = 0$, this equation either has $0$ or $2$ real roots, according to whether $256 a c^3 - 27 b^4$ is positive or negative, respectively. (If the discriminant is zero, either is possible.)
– Travis
Nov 15 at 12:28
2
At any rate, the general formula is somewhat unpleasant, and probably for your (empirical) setting a numerical solution is easier and more reasonable besides.
– Travis
Nov 15 at 12:29
add a comment |
2
this link may be helpful: math.stackexchange.com/questions/785/…
– Sujit Bhattacharyya
Nov 15 at 11:52
Computing the discriminant and using Descartes' Rule of Signs shows that unless $c = 0$, this equation either has $0$ or $2$ real roots, according to whether $256 a c^3 - 27 b^4$ is positive or negative, respectively. (If the discriminant is zero, either is possible.)
– Travis
Nov 15 at 12:28
2
At any rate, the general formula is somewhat unpleasant, and probably for your (empirical) setting a numerical solution is easier and more reasonable besides.
– Travis
Nov 15 at 12:29
2
2
this link may be helpful: math.stackexchange.com/questions/785/…
– Sujit Bhattacharyya
Nov 15 at 11:52
this link may be helpful: math.stackexchange.com/questions/785/…
– Sujit Bhattacharyya
Nov 15 at 11:52
Computing the discriminant and using Descartes' Rule of Signs shows that unless $c = 0$, this equation either has $0$ or $2$ real roots, according to whether $256 a c^3 - 27 b^4$ is positive or negative, respectively. (If the discriminant is zero, either is possible.)
– Travis
Nov 15 at 12:28
Computing the discriminant and using Descartes' Rule of Signs shows that unless $c = 0$, this equation either has $0$ or $2$ real roots, according to whether $256 a c^3 - 27 b^4$ is positive or negative, respectively. (If the discriminant is zero, either is possible.)
– Travis
Nov 15 at 12:28
2
2
At any rate, the general formula is somewhat unpleasant, and probably for your (empirical) setting a numerical solution is easier and more reasonable besides.
– Travis
Nov 15 at 12:29
At any rate, the general formula is somewhat unpleasant, and probably for your (empirical) setting a numerical solution is easier and more reasonable besides.
– Travis
Nov 15 at 12:29
add a comment |
active
oldest
votes
active
oldest
votes
active
oldest
votes
active
oldest
votes
active
oldest
votes
Thanks for contributing an answer to Mathematics Stack Exchange!
- Please be sure to answer the question. Provide details and share your research!
But avoid …
- Asking for help, clarification, or responding to other answers.
- Making statements based on opinion; back them up with references or personal experience.
Use MathJax to format equations. MathJax reference.
To learn more, see our tips on writing great answers.
Some of your past answers have not been well-received, and you're in danger of being blocked from answering.
Please pay close attention to the following guidance:
- Please be sure to answer the question. Provide details and share your research!
But avoid …
- Asking for help, clarification, or responding to other answers.
- Making statements based on opinion; back them up with references or personal experience.
To learn more, see our tips on writing great answers.
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%2f2999602%2fsolving-a-x4-b-x-c-0%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
2
this link may be helpful: math.stackexchange.com/questions/785/…
– Sujit Bhattacharyya
Nov 15 at 11:52
Computing the discriminant and using Descartes' Rule of Signs shows that unless $c = 0$, this equation either has $0$ or $2$ real roots, according to whether $256 a c^3 - 27 b^4$ is positive or negative, respectively. (If the discriminant is zero, either is possible.)
– Travis
Nov 15 at 12:28
2
At any rate, the general formula is somewhat unpleasant, and probably for your (empirical) setting a numerical solution is easier and more reasonable besides.
– Travis
Nov 15 at 12:29