Question

w^4-14w^2-2=0
solve

Answer 1

gimmick is to put
\[x=w^2\] and write
\[x^2-14x-2=0\] solve for x

Answer 2

then go back.
solve for x via
\[x^2-14x=2\]
\[(x-7)^2=2+49\]
\[(x-7)^2=51\]
\[x-7=\pm\sqrt{51}\]
\[x=7\pm\sqrt{51}\]

Answer 3

now
\[7-\sqrt{51}<0\] so you cannot have
\[w^2=7-\sqrt{51}\] with real numbers. therefore
\[w^2=7+\sqrt{51}\] and finally
\[w=\pm\sqrt{7+\sqrt{51}}\]

Answer 4

thank you!

Answer 5

yw