Solving Fourth Degree Equation: Get Answers Now

[mathfriends]

[URL]http://www13.0zz0.com/2011/07/15/22/754178270.jpg[/URL]

I Need help solving this fourth degree equation

 

[Pengwuino]

"I want answers" is not how you're going to get results. Read the rules of the forums.

 

[micromass]

Read: http://en.wikipedia.org/wiki/Quartic_equation

 

[pmsrw3]

I want answers

Me, too.

 

[Dr. Seafood]

lol comic sans

 

[mathfriends]

thank you

Finally I found a solution

 

[I like Serena]

Hmm.
I understand the solution x=-1.
But how did you get the other solutions from $9x^3-4x+1=0$?

 

[HallsofIvy]

The simplest way to approach this problem is to use the "rational root theorem".

If r is a rational number satisfying $a_nx^n+ a_{n-1}x^{n-1}+ \cdot\cdot\cdot+ a_1x+ a_0= 0$ where the coefficients are all integers, then r= p/q where p is an integer evenly dividing $a_0$ and q is an integer evenly dividing $a_n$.

Of course, it is not necessary than such an equation have any rational roots but it is worth trying. Here, the leading coefficient, $a_n$, is 9, which has factors $\pm 1, \pm 3, \pm 9$ and the constant term, $a_0$ is 1, which has factors $\pm 1$ so the only possible rational roots are $\pm 1, \pm 1/3, \pm 1/9$.

Putting those into the equation, we see that if
$$9(1)^4+ 9(1)^3- 4(1)^2- 3(1)+ 1= 19- 7= 12\ne 0$$
$$9(-1)^4+ 9(-1)^3- 4(-1)^2- 3(-1)+ 1= -9+ 9- 4+ 3+ 1= -1+1= 0$$
$$9(1/3)^4+ 9(1/3)^3- 4(1/3)^2- 3(1/3)+ 1= 1/9+ 1/3- 4/9- 1+ 1= 1/3- 1/3- 1+ 1= 0$$
We can stop here. Seeing that x= -1 and x= 1/3 are roots, we can divide by x+ 1 and x- 1/3 (not "0.3333") to get a quadratic equation that we can solve using the quadratic formula.

(There is a "quartic formula", http://www.sosmath.com/algebra/factor/fac12/fac12.html, but it is extermely complicated.)

 

[I like Serena]

The simplest way to approach this problem is to use the "rational root theorem".

Nice! I didn't know this one yet!

 

[mathfriends]

this is another solution

Is this true ?

 

[Redbelly98]

Moderator's note: thread moved from "General Math".

Please help us moderate the forum by reporting misplaced homework posts.

Thank you.

 

[Redbelly98]

this is another solution

Is this true ?

Why don't you tell us? Are those answers the same as the ones you already know:

thank you

Finally I found a solution

 

[mathfriends]

Why don't you tell us? Are those answers the same as the ones you already know:

yeah,I just want clarification

 

[I like Serena]

this is another solution

Is this true ?

What do you think?
Can you think of a way to derive this solution?

Btw, are you familiar with solving a quadratic equation?
That is, an equation of the form ax² + bx + c = 0.

 

[mathfriends]

What do you think?
Can you think of a way to derive this solution?

Btw, are you familiar with solving a quadratic equation?
That is, an equation of the form ax² + bx + c = 0.

Yes, do you mean you want the law of quadratic equation

 

[I like Serena]

What I would like is for you to give us some insight into what you are thinking.
That would make it easier for us to help you.

As it is we have no clue how you got this problem or why you want to solve it.
I'm assuming you're supposed to learn how to solve such equations.
At least that's what we're trying to help you with.

How did you get this problem?
And what is your purpose with it?

 

[mathfriends]

thank you ( i like serena ) for help me .

 

[keyfob]

The simplest way to approach this problem is to use the "rational root theorem".

If r is a rational number satisfying $a_nx^n+ a_{n-1}x^{n-1}+ \cdot\cdot\cdot+ a_1x+ a_0= 0$ where the coefficients are all integers, then r= p/q where p is an integer evenly dividing $a_0$ and q is an integer evenly dividing $a_n$.

Of course, it is not necessary than such an equation have any rational roots but it is worth trying. Here, the leading coefficient, $a_n$, is 9, which has factors $\pm 1, \pm 3, \pm 9$ and the constant term, $a_0$ is 1, which has factors $\pm 1$ so the only possible rational roots are $\pm 1, \pm 1/3, \pm 1/9$.

Putting those into the equation, we see that if
$$9(1)^4+ 9(1)^3- 4(1)^2- 3(1)+ 1= 19- 7= 12\ne 0$$
$$9(-1)^4+ 9(-1)^3- 4(-1)^2- 3(-1)+ 1= -9+ 9- 4+ 3+ 1= -1+1= 0$$
$$9(1/3)^4+ 9(1/3)^3- 4(1/3)^2- 3(1/3)+ 1= 1/9+ 1/3- 4/9- 1+ 1= 1/3- 1/3- 1+ 1= 0$$
We can stop here. Seeing that x= -1 and x= 1/3 are roots, we can divide by x+ 1 and x- 1/3 (not "0.3333") to get a quadratic equation that we can solve using the quadratic formula.

(There is a "quartic formula", http://www.sosmath.com/algebra/factor/fac12/fac12.html, but it is extermely complicated.)

Dude! This is an awesome theorem, thanks for that.