Solve x^4 - x - 2 = 0 (How to get Wolfram's answer?)

[Stallmp]

Hello,

Solve x^4-x-2=0

According to Wolfram Alpha, the answer is

x=-1
x=\(\displaystyle \dfrac{1}{6}\, \left(\, 2\, -\, 4\, \sqrt[3]{\strut \dfrac{2}{47\, +\, 3\sqrt{249\,}}\,}\, +\, 2^{\frac{2}{3}}\, \sqrt[3]{\strut 47\, +\, 3\sqrt{249\,}\,}\, \right)\)

\(\displaystyle x\, \approx\, -0.17660\, \pm\, 1.20282i\)

But could anyone show me step by step how to solve this problem?

 

[Deleted member 4993]

Hello,

According to Wolfram Alpha, the answer is

x=-1
x=

But could anyone show me step by step how to solve this problem?

There are two more (complex) solutions. What method/s have been taught to you in class?

 

[Stallmp]

What do you mean with complex solutions ? I haven't heard of that before. I have been taught the ABC Formula and P-substitutions (and definitely more techniques but I think these are the most important ones). But I was wondering what the steps were to get the 2 (basic) solutions from Wolfram.

 

[tkhunny]

\(\displaystyle x^{2} - 1 = 0\) has two Real Solutions. Can you name them?

\(\displaystyle x^{2} + 1 = 0\) has two Complex Solution. Can you name them?

 

[Stallmp]

\(\displaystyle x^{2} - 1 = 0\) has two Real Solutions. Can you name them?

\(\displaystyle x^{2} + 1 = 0\) has two Complex Solution. Can you name them?

Is the real solution: x=1 and x=-1?
I still don't know what complex solutions are. I think it has to do with i? But could you guys explain the real solution to me step by step?

 

[stapel]

Solve x^4-x-2=0

According to Wolfram Alpha, the answer is

x=-1
x=\(\displaystyle \dfrac{1}{6}\, \left(\, 2\, -\, 4\, \sqrt[3]{\strut \dfrac{2}{47\, +\, 3\sqrt{249\,}}\,}\, +\, 2^{\frac{2}{3}}\, \sqrt[3]{\strut 47\, +\, 3\sqrt{249\,}\,}\, \right)\)

\(\displaystyle x\, \approx\, -0.17660\, \pm\, 1.20282i\)

But could anyone show me step by step how to solve this problem?

Ouch! They were supposed to have covered this material in class before assigning you homework on it!

To learn how to solve general polynomials (to the extent that they can be solved algebraically), try here. Study at least three lessons from the listing. Then:

1) I think some of the other replies here have misread the original equation, because what you posted does not have a factor of x^2 + 1 or x^2 - 1.

2) Do a quick graph of the related polynomial function, f(x) = (what you posted). The zeroes that you're seeking are the (real-valued) x-intercepts.

3) Use synthetic division to quickly prove the one zero that's obvious from the graph.

4) The other real-valued zero is, as you've seen on Wolfram's site, not "nice". To obtain this result, you'd have to try delving into solving the cubic that's left after you factor out the one easy value. Nobody does this, because the results are, well, exactly as messy and painful as you've seen.

5) I don't know how you're supposed to obtain the complex-valued results, especially since your class hasn't even covered how to obtain the real-valued results.

You might want to have a talk with the instructor. Maybe there's a typo in the exercise...? Thank you!

 

[Alexander123]

Hello,

Solve x^4-x-2=0

Hi, the method I used to solve this equation to get "x" to equal to "-1" was by using synthetic division to find the zero,or zero's of this equation. -1 is indeed, correct and the way I started was this.


First, find the possible zero of this equation by using the zero test formula. You use the constant which is "-2" and find the factors of -2 which will be your possible zeros.

-2 is; 1,-1,2,-2. Binomial form are; (x-1) for 1,(x+1) for -1,(x-2) for 2,and (x+2) for -2.

Now, use synthetic division and you'll find out that -1 is the only number that makes this equation equal to zero.

Let me know if you need help with synthetic division.

 

[mmm4444bot]

First, find the possible zero of this equation by using the zero test formula. You use the constant which is "-2" and find the factors of -2 which will be your possible zeros.

This works, if the leading coefficient is 1, -1, 2, or -2. That's a "special case" of the Rational Root Theorem.

If the leading coefficient is any other value, then you would need to consider factors of the leading coefficient, also.

Possible Rational roots would be formed by taking ratios of the factors, where the numerator is a factor of the constant and the denominator is a factor of the leading coefficient. :cool:

 

[mmm4444bot]

I was wondering what the steps were to get the 2 [Real] solutions from Wolfram.

Post #7 gave you "steps" for getting x=-1 (using the Rational Root Theorem and testing candidates).

I'll give you the steps (one at a time) for the other solution, but you'll need to do the work.

Be forewarned that this method involves changing coordinates a few times (from x to y to z to u, to form a quadratic equation to solve for u) and then back-substituting (from u to z to y to x) to obtain the final result. You will need a few hours, for all of the algebraic manipulations involved.

Here we go...

STEP 1 (You've already done it!)

Factor the left-hand side of x^4 - x - 2 = 0

So, either

x + 1 = 0

OR

x^3 - x^2 + x - 2 = 0

STEP 2

Begin solving x^3 - x^2 + x - 2 = 0 by making the following substitution:

x = y + 1/3

and then multiplying out the left-hand side, to combine like terms

Show me what you get :cool:

 

[mmm4444bot]

I still don't know what complex solutions are. I think it has to do with i?

Yes, they do.

i is the square root of -1.

In other words, i^2 = -1.

The number i is imaginary, in the sense that it does not belong to the set of Real numbers. Mathematicians defined it, in order to completely solve certain equations.

By adding products of i to Real numbers, we get the set of Complex numbers.

A Complex number takes the form a + b*i, where a and b are any Real numbers.

So there's your introduction to Complex numbers.

You'll need to use them, if you want to find the solutions of equations like

x^2 + x + 1 = 0

because the solutions are

x = -1/2 + sqrt(3)/2 * i

OR

x = -1/2 - sqrt(3)/2 * i