SOLUTION: X^4+x^2+1=0

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Question 846997: X^4+x^2+1=0
Found 2 solutions by lambo, josh_jordan:
Answer by lambo(6) About Me  (Show Source):
You can put this solution on YOUR website!
X^4+x^2+1=0
X^6+1=0
-1 -1
X^6=-1
-1^6=-1
X=-1

Answer by josh_jordan(263) About Me  (Show Source):
You can put this solution on YOUR website!
This is quite the complex problem! So, let's get started.

If we rewrite x%5E4%2Bx%5E2%2B1=0 in the form of a quadratic equation, we will be able to find our zeroes. We can do this by rewriting as:

%28x%5E2%29%5E2%2B%28x%29x%2B1=0

This puts our equation in standard quadratic form of ax^2+bx+c=0. Now, we can replace the first x inside the parenthesis with a letter and replace the second x on the outside of the parenthesis with the same letter. Let's use u:

u%5E2%2Bu%2B1=0

Now we can solve using the quadratic formula. This will give us:

u=%28-1+%2B-+sqrt%28+-3+%29%29%2F%282%29+

Now, here comes the tricky part. We have to replace our u with our original variable and power: x^2, which will turn this into:

x%5E2=%28-1+%2B-+sqrt%28+-3+%29%29%2F%282%29+

Because we want the value of x and not x^2, we will have to take the square root of both sides. Remember, when we take the square root of both sides of an equation, we have to add the +- sign in front. Doing this is going to give us 4 zeros:

x=+%2B-+%28sqrt%28%28-1+%2B-+sqrt%28+-3+%29%29%2F%282%29%29%29

Next, you will want to convert the square root of -3 into an imaginary:

x=+%2B-+%28sqrt%28%28-1+%2B-+sqrt%28+3+%29%2Ai%29%2F%282%29%29%29

Finally, you will need to rationalize the denominator since the denominator of the fraction is the square root of 2. This will give you your final answer:

x=+%2B-+%28sqrt%28%282%28-1+%2B-+sqrt%28+3+%29%2Ai%29%29%29%2F%282%29%29

For some reason, the answer typed out the word "PLUS_MINNUS-" instead of "+-". Just replace the "PLUS_MINNUS-" with the plus/minus symbol (+-).
NOTE: You will notice the numerator of the fraction contains a square root inside of a square root (nested root). You will more than likely not be required to denest this root, so it was not denested in this solution.