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

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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

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nothing wrong at the beginning. you get \[w^2=2+\sqrt{6}\] \[w=\pm \sqrt{2+\sqrt{6}}\]
it is your final calculation that is off. you only need to take the square root of your solution for \[w^2\]
now \[2-\sqrt{6}<0\] so if you are working only with real numbers this answer gives no solution for w

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did this make sense? the only mistake you made was taking the square root of \[2+\sqrt{6}\]
shes an idiot
thats whats off
you have to put the square root sign over the whole expression
w=sqrt [ 2+sqrt [6] ]

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