## anonymous 5 years ago What is the solution for w: w^4-18w^2-2=0

1. anonymous
2. anonymous

yeah, it looks like that one would be best done with subbing in x for w^2, and then using the quadratic on it, then subbing back, and you might get something pretty ugly

3. anonymous

wolfram alpha should be able to solve it too but i would just use it to compare your results

4. anonymous

I look at this and it is like Greek to me..I mean I kinda get it, this is where I get confused...

5. anonymous

Alright you take the w^2's and replace them with x's. So you will get x^2-18x-2=0. Now that you have it in the quadratic form which is ax^2+bx+c=0. You can determine the values of a, b, and c. Thus a = 1, b =-18, and c =-2. With this in mind you apply the quardratic formula which is...$-b \pm \sqrt{b ^{2}-4ac}\div 2a$ you simply will now plug in the values of a,b,c that was listed above. You will get 2 answers.. one using the negative -b- and one using the positive -b+. And your done!