What is the solution for w: w^4-18w^2-2=0
http://www.algebra.com/algebra/homework/quadratic/Quadratic_Equations.faq.question.332342.html
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
wolfram alpha should be able to solve it too but i would just use it to compare your results
I look at this and it is like Greek to me..I mean I kinda get it, this is where I get confused...
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!
Join our real-time social learning platform and learn together with your friends!