Question

Need help understanding this problem and how to get this answer. Please explain. Quadratic Equations

w^4-4w^2-2=0

Answer:
sqrt{2+sqrt{6}}, sqrt{2-sqrt{6}
-sqrt{2+sqrt{6}}, -sqrt{2-sqrt{6}}

Answer 1

\[\sqrt{2+\sqrt{6}} , \sqrt{2-\sqrt{6}}\]

Answer 2

\[-\sqrt{2+\sqrt{6}} , -2\sqrt{-\sqrt{6}}\]

Answer 3

2 should be inside the radical my bad

Answer 4

I think we might want to say:
\[w^2=x\]

Answer 5

yea

Answer 6

Then our equation becomes:
\[x^2-4x-2=0\]

Answer 7

my book says that w^2 = u

Answer 8

And we can use the quadratic formula on that.

Answer 9

Ok. Then use u

Answer 10

Then when we find the two values of u, we can replace u with w^2 and solve those two equations.

Answer 11

im still lost, i understand up to that but than they get these radicals out of no where. What do i multiply add divide or what not

Answer 12

do I use -b +/- √b^2-4ac/2a

Answer 13

to get to that answer, unfortunately my book doesnt explain very well

Answer 14

\[u=\frac{4\pm \sqrt{16-4(-2)}}{2}=\frac{4\pm \sqrt{24}}{2}=\frac{4\pm2\sqrt{6}}{2}=2\pm \sqrt{6}\]

Answer 15

But u = w^2

Answer 16

\[w^2=2+\sqrt{6}\]

Answer 17

ah ok. I see where i went wrong. My book just skipped that whole step loll was confused that they were still referring to previous chapters of work

Answer 18

\[w=\pm \sqrt{2+\sqrt{6}}\]

Answer 19

yea I see now thanks a lot. Was just lost

Answer 20

yw