SOLUTION: Solve. w^4-12w^2-2=0 I did a similar problem and was supposed to come up with 4 different radicals and I only came up with 2 for my answer and was told it was wrong.

Algebra ->  Quadratic Equations and Parabolas  -> Quadratic Equations Lessons  -> Quadratic Equation Lesson -> SOLUTION: Solve. w^4-12w^2-2=0 I did a similar problem and was supposed to come up with 4 different radicals and I only came up with 2 for my answer and was told it was wrong.      Log On


   

Question 130949: Solve.
w^4-12w^2-2=0
I did a similar problem and was supposed to come up with 4 different radicals and I only came up with 2 for my answer and was told it was wrong.
Answer by solver91311(24713) About Me  (Show Source):
You can put this solution on YOUR website!
w%5E4-12w%5E2-2=0

Let u=w%5E2, so:

u%5E2-12u-2=0

Complete the square:
u%5E2-12u=2

u%5E2-12u%2B36=38

%28u-6%29%5E2=38

u-6=sqrt%2838%29 or u-6=-sqrt%2838%29

u%5B1%5D=6%2Bsqrt%2838%29 or u%5B2%5D=6-sqrt%2838%29

But u=w%5E2, so w=sqrt%28u%29 or w=-sqrt%28u%29

w=sqrt%28u%5B1%5D%29=sqrt%286%2Bsqrt%2838%29%29, or

w=sqrt%28u%5B2%5D%29=sqrt%286-sqrt%2838%29%29, or

w=-sqrt%28u%5B1%5D%29=-sqrt%286%2Bsqrt%2838%29%29, or

w=-sqrt%28u%5B2%5D%29=-sqrt%286-sqrt%2838%29%29

Anytime you have a 4th degree polynomial equation, you must expect to have exactly 4 roots. They may be 4 real roots, 2 pairs of complex congujate roots, or 2 real roots and 1 pair of complex conjugate roots. Pairs of real roots might be identical (multiplicity greater than 1) but there will ALWAYS be 4 roots when the highest degree term is 4.