anonymous
  • anonymous
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}}
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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jamiebookeater
  • jamiebookeater
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anonymous
  • anonymous
\[\sqrt{2+\sqrt{6}} , \sqrt{2-\sqrt{6}}\]
anonymous
  • anonymous
\[-\sqrt{2+\sqrt{6}} , -2\sqrt{-\sqrt{6}}\]
anonymous
  • anonymous
2 should be inside the radical my bad

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Mertsj
  • Mertsj
I think we might want to say: \[w^2=x\]
anonymous
  • anonymous
yea
Mertsj
  • Mertsj
Then our equation becomes: \[x^2-4x-2=0\]
anonymous
  • anonymous
my book says that w^2 = u
Mertsj
  • Mertsj
And we can use the quadratic formula on that.
Mertsj
  • Mertsj
Ok. Then use u
Mertsj
  • Mertsj
Then when we find the two values of u, we can replace u with w^2 and solve those two equations.
anonymous
  • anonymous
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
anonymous
  • anonymous
do I use -b +/- √b^2-4ac/2a
anonymous
  • anonymous
to get to that answer, unfortunately my book doesnt explain very well
Mertsj
  • Mertsj
\[u=\frac{4\pm \sqrt{16-4(-2)}}{2}=\frac{4\pm \sqrt{24}}{2}=\frac{4\pm2\sqrt{6}}{2}=2\pm \sqrt{6}\]
Mertsj
  • Mertsj
But u = w^2
Mertsj
  • Mertsj
\[w^2=2+\sqrt{6}\]
anonymous
  • anonymous
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
Mertsj
  • Mertsj
\[w=\pm \sqrt{2+\sqrt{6}}\]
anonymous
  • anonymous
yea I see now thanks a lot. Was just lost
Mertsj
  • Mertsj
yw

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