SOLUTION: I need help solving problem What is the solution for this problem? w^4-2w^2-2=0

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Question 407266: I need help solving problem
What is the solution for this problem?
w^4-2w^2-2=0
Answer by nerdybill(7384) About Me  (Show Source):
You can put this solution on YOUR website!
w^4-2w^2-2=0
.
Start by:
Letting x = w^2
then we can rewrite as:
x^2-2x-2=0
this is a quadratic that is not factorable so we apply the quadratic formula to get our solution of:
x = {2.732, -0.732}
we can toss out the negative solution (extraneous) to get:
x = 2.732
.
But, we are looking for w:
x = w^2
2.732 = w^2
1.653 = w
.
details of quadratic follows:
Solved by pluggable solver: SOLVE quadratic equation with variable
Quadratic equation ax%5E2%2Bbx%2Bc=0 (in our case 1x%5E2%2B-2x%2B-2+=+0) has the following solutons:

x%5B12%5D+=+%28b%2B-sqrt%28+b%5E2-4ac+%29%29%2F2%5Ca

For these solutions to exist, the discriminant b%5E2-4ac should not be a negative number.

First, we need to compute the discriminant b%5E2-4ac: b%5E2-4ac=%28-2%29%5E2-4%2A1%2A-2=12.

Discriminant d=12 is greater than zero. That means that there are two solutions: +x%5B12%5D+=+%28--2%2B-sqrt%28+12+%29%29%2F2%5Ca.

x%5B1%5D+=+%28-%28-2%29%2Bsqrt%28+12+%29%29%2F2%5C1+=+2.73205080756888
x%5B2%5D+=+%28-%28-2%29-sqrt%28+12+%29%29%2F2%5C1+=+-0.732050807568877

Quadratic expression 1x%5E2%2B-2x%2B-2 can be factored:
1x%5E2%2B-2x%2B-2+=+1%28x-2.73205080756888%29%2A%28x--0.732050807568877%29
Again, the answer is: 2.73205080756888, -0.732050807568877. Here's your graph:
graph%28+500%2C+500%2C+-10%2C+10%2C+-20%2C+20%2C+1%2Ax%5E2%2B-2%2Ax%2B-2+%29