solve
w^4-4w²-2=0
cant figure out what i am doing wrong
u=w²
u²-4u-2=0
u=2-(sqrt)6 or u=2+(sqrt)6
w²=2-(sqrt)6 or w²=2+(sqrt)6
w=(sqrt)2+(sqrt)6, -(sqrt)2+(sqrt)6
tell me what i am doing wrong and how to fix it plz

Question

solve 
w^4-4w²-2=0 
cant figure out what i am doing wrong 
u=w² 
u²-4u-2=0 
u=2-(sqrt)6 or u=2+(sqrt)6
 w²=2-(sqrt)6 or w²=2+(sqrt)6 
w=(sqrt)2+(sqrt)6, -(sqrt)2+(sqrt)6
 tell me what i am doing wrong and how to fix it plz

anonymous · Answer

nothing wrong at the beginning.  you get 
$$w^2=2+\sqrt{6}$$
$$w=\pm \sqrt{2+\sqrt{6}}$$

anonymous · Answer

it is your final calculation that is off.  you only need to take the square root of your solution for $$w^2$$

anonymous · Answer

now 
$$2-\sqrt{6}<0$$ so if you are working only with real numbers this answer gives no solution for w

anonymous · Answer

did this make sense?  the only mistake you made was taking the square root of $$2+\sqrt{6}$$

anonymous · Answer

shes an idiot

anonymous · Answer

thats whats off

anonymous · Answer

you have to put the square root sign over the whole expression

anonymous · Answer

w=sqrt [ 2+sqrt [6]  ]