Solving an equation containing 4th power of variable.
I know how to solve Quadratic equations. Recently i came across the
equation of type $ax^4 + bx^2 + c = 0$ and i had to solve it. So what i
did is that i supposed $x^2 = y$ so that the above equation becomes:
$ay^2 + by + c = 0$ and i found y by quadratic formula and took the
square root of y getting x.I.e:
$x = sqrt{-b pm frac{sqrt{b^2 - 4ac}}{2a}}$
But i have to confirm whether this solution is correct or not. If not then what is the true solution.
quadratics
edited Aug 29 '15 at 14:34
Oussama Boussif
2,887722
asked Aug 29 '15 at 14:21
user258250user258250
24518
|
show 2 more comments
I know how to solve Quadratic equations. Recently i came across the
equation of type $ax^4 + bx^2 + c = 0$ and i had to solve it. So what i
did is that i supposed $x^2 = y$ so that the above equation becomes:
$ay^2 + by + c = 0$ and i found y by quadratic formula and took the
square root of y getting x.I.e:
$x = sqrt{-b pm frac{sqrt{b^2 - 4ac}}{2a}}$
But i have to confirm whether this solution is correct or not. If not then what is the true solution.
quadratics
edited Aug 29 '15 at 14:34
Oussama Boussif
2,887722
asked Aug 29 '15 at 14:21
user258250user258250
24518
1
$x=pmsqrt{frac{-bpm sqrt{b^2-4ac}}{2a}}$.It is the correct solution since quadratic formulae must work but may be complex.
– kingW3
Aug 29 '15 at 14:24
This method will work well on this type of equation. You can plug in your final answers into the original equation to check that they're solutions. However, since you will get four solutions out of this method (and all solutions will arise in this way), you are safe knowing that these are all of the solutions.
– Michael Burr
Aug 29 '15 at 14:24
Side note: but when you have equation of the form $ax^4+bx^3+cx^2+dx+e$, you can use Ferrar's formula. Look here: en.wikipedia.org/wiki/Quartic_function.
– VanDerWarden
Aug 29 '15 at 14:26
It is right! (y)
– Aditya Agarwal
Aug 29 '15 at 14:32
Often, people just leave it as 'a solution of $ax^4+bx^3+cx^2+dx+e=0$'. There are other methods (for example a method called 'Newton's Method') that give the answer to as many decimal places as you want.
– Empy2
Aug 29 '15 at 14:34
|
show 2 more comments
I know how to solve Quadratic equations. Recently i came across the
equation of type $ax^4 + bx^2 + c = 0$ and i had to solve it. So what i
did is that i supposed $x^2 = y$ so that the above equation becomes:
$ay^2 + by + c = 0$ and i found y by quadratic formula and took the
square root of y getting x.I.e:
$x = sqrt{-b pm frac{sqrt{b^2 - 4ac}}{2a}}$
But i have to confirm whether this solution is correct or not. If not then what is the true solution.
quadratics
edited Aug 29 '15 at 14:34
Oussama Boussif
2,887722
asked Aug 29 '15 at 14:21
user258250user258250
24518
I know how to solve Quadratic equations. Recently i came across the
equation of type $ax^4 + bx^2 + c = 0$ and i had to solve it. So what i
did is that i supposed $x^2 = y$ so that the above equation becomes:
$ay^2 + by + c = 0$ and i found y by quadratic formula and took the
square root of y getting x.I.e:
$x = sqrt{-b pm frac{sqrt{b^2 - 4ac}}{2a}}$
But i have to confirm whether this solution is correct or not. If not then what is the true solution.
quadratics
quadratics
edited Aug 29 '15 at 14:34
Oussama Boussif
2,887722
asked Aug 29 '15 at 14:21
user258250user258250
24518
edited Aug 29 '15 at 14:34
Oussama Boussif
2,887722
asked Aug 29 '15 at 14:21
user258250user258250
24518
edited Aug 29 '15 at 14:34
Oussama Boussif
2,887722
edited Aug 29 '15 at 14:34
Oussama Boussif
2,887722
edited Aug 29 '15 at 14:34
Oussama Boussif
2,887722
2,887722
asked Aug 29 '15 at 14:21
user258250user258250
24518
asked Aug 29 '15 at 14:21
user258250user258250
24518
asked Aug 29 '15 at 14:21
user258250user258250
24518
24518
1
$x=pmsqrt{frac{-bpm sqrt{b^2-4ac}}{2a}}$.It is the correct solution since quadratic formulae must work but may be complex.
– kingW3
Aug 29 '15 at 14:24
This method will work well on this type of equation. You can plug in your final answers into the original equation to check that they're solutions. However, since you will get four solutions out of this method (and all solutions will arise in this way), you are safe knowing that these are all of the solutions.
– Michael Burr
Aug 29 '15 at 14:24
Side note: but when you have equation of the form $ax^4+bx^3+cx^2+dx+e$, you can use Ferrar's formula. Look here: en.wikipedia.org/wiki/Quartic_function.
– VanDerWarden
Aug 29 '15 at 14:26
It is right! (y)
– Aditya Agarwal
Aug 29 '15 at 14:32
Often, people just leave it as 'a solution of $ax^4+bx^3+cx^2+dx+e=0$'. There are other methods (for example a method called 'Newton's Method') that give the answer to as many decimal places as you want.
– Empy2
Aug 29 '15 at 14:34
|
show 2 more comments
1
$x=pmsqrt{frac{-bpm sqrt{b^2-4ac}}{2a}}$.It is the correct solution since quadratic formulae must work but may be complex.
– kingW3
Aug 29 '15 at 14:24
This method will work well on this type of equation. You can plug in your final answers into the original equation to check that they're solutions. However, since you will get four solutions out of this method (and all solutions will arise in this way), you are safe knowing that these are all of the solutions.
– Michael Burr
Aug 29 '15 at 14:24
Side note: but when you have equation of the form $ax^4+bx^3+cx^2+dx+e$, you can use Ferrar's formula. Look here: en.wikipedia.org/wiki/Quartic_function.
– VanDerWarden
Aug 29 '15 at 14:26
It is right! (y)
– Aditya Agarwal
Aug 29 '15 at 14:32
Often, people just leave it as 'a solution of $ax^4+bx^3+cx^2+dx+e=0$'. There are other methods (for example a method called 'Newton's Method') that give the answer to as many decimal places as you want.
– Empy2
Aug 29 '15 at 14:34
1
1
$x=pmsqrt{frac{-bpm sqrt{b^2-4ac}}{2a}}$.It is the correct solution since quadratic formulae must work but may be complex.
– kingW3
Aug 29 '15 at 14:24
$x=pmsqrt{frac{-bpm sqrt{b^2-4ac}}{2a}}$.It is the correct solution since quadratic formulae must work but may be complex.
– kingW3
Aug 29 '15 at 14:24
This method will work well on this type of equation. You can plug in your final answers into the original equation to check that they're solutions. However, since you will get four solutions out of this method (and all solutions will arise in this way), you are safe knowing that these are all of the solutions.
– Michael Burr
Aug 29 '15 at 14:24
This method will work well on this type of equation. You can plug in your final answers into the original equation to check that they're solutions. However, since you will get four solutions out of this method (and all solutions will arise in this way), you are safe knowing that these are all of the solutions.
– Michael Burr
Aug 29 '15 at 14:24
Side note: but when you have equation of the form $ax^4+bx^3+cx^2+dx+e$, you can use Ferrar's formula. Look here: en.wikipedia.org/wiki/Quartic_function.
– VanDerWarden
Aug 29 '15 at 14:26
Side note: but when you have equation of the form $ax^4+bx^3+cx^2+dx+e$, you can use Ferrar's formula. Look here: en.wikipedia.org/wiki/Quartic_function.
– VanDerWarden
Aug 29 '15 at 14:26
It is right! (y)
– Aditya Agarwal
Aug 29 '15 at 14:32
It is right! (y)
– Aditya Agarwal
Aug 29 '15 at 14:32
Often, people just leave it as 'a solution of $ax^4+bx^3+cx^2+dx+e=0$'. There are other methods (for example a method called 'Newton's Method') that give the answer to as many decimal places as you want.
– Empy2
Aug 29 '15 at 14:34
Often, people just leave it as 'a solution of $ax^4+bx^3+cx^2+dx+e=0$'. There are other methods (for example a method called 'Newton's Method') that give the answer to as many decimal places as you want.
– Empy2
Aug 29 '15 at 14:34
|
show 2 more comments
1 Answer
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Yes, quite correct, except that the outer square-root has a separate $pm$. So $x=pmsqrt{-bpmsqrt{b^2-4ac}/2a}$
answered Aug 29 '15 at 14:24
Empy2Empy2
33.5k12261
add a comment |
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Yes, quite correct, except that the outer square-root has a separate $pm$. So $x=pmsqrt{-bpmsqrt{b^2-4ac}/2a}$
answered Aug 29 '15 at 14:24
Empy2Empy2
33.5k12261
add a comment |
Yes, quite correct, except that the outer square-root has a separate $pm$. So $x=pmsqrt{-bpmsqrt{b^2-4ac}/2a}$
answered Aug 29 '15 at 14:24
Empy2Empy2
33.5k12261
add a comment |
Yes, quite correct, except that the outer square-root has a separate $pm$. So $x=pmsqrt{-bpmsqrt{b^2-4ac}/2a}$
answered Aug 29 '15 at 14:24
Empy2Empy2
33.5k12261
Yes, quite correct, except that the outer square-root has a separate $pm$. So $x=pmsqrt{-bpmsqrt{b^2-4ac}/2a}$
answered Aug 29 '15 at 14:24
Empy2Empy2
33.5k12261
answered Aug 29 '15 at 14:24
Empy2Empy2
33.5k12261
answered Aug 29 '15 at 14:24
Empy2Empy2
33.5k12261
answered Aug 29 '15 at 14:24
Empy2Empy2
33.5k12261
33.5k12261
add a comment |
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1
$x=pmsqrt{frac{-bpm sqrt{b^2-4ac}}{2a}}$.It is the correct solution since quadratic formulae must work but may be complex.
– kingW3
Aug 29 '15 at 14:24
This method will work well on this type of equation. You can plug in your final answers into the original equation to check that they're solutions. However, since you will get four solutions out of this method (and all solutions will arise in this way), you are safe knowing that these are all of the solutions.
– Michael Burr
Aug 29 '15 at 14:24
Side note: but when you have equation of the form $ax^4+bx^3+cx^2+dx+e$, you can use Ferrar's formula. Look here: en.wikipedia.org/wiki/Quartic_function.
– VanDerWarden
Aug 29 '15 at 14:26
It is right! (y)
– Aditya Agarwal
Aug 29 '15 at 14:32
Often, people just leave it as 'a solution of $ax^4+bx^3+cx^2+dx+e=0$'. There are other methods (for example a method called 'Newton's Method') that give the answer to as many decimal places as you want.
– Empy2
Aug 29 '15 at 14:34