Question

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

Answer 1

factored or solved?

Answer 2

solved last please using radical expression

Answer 3

x = +1,-1,+2,-2

Answer 4

you have to factoize it to get 2 quadratic factors each of which gives you 2 roots

Answer 5

that is not right

Answer 6

Let n = w^2

Then substituting for w^2:

\[n ^{2}-2n+2 = 0\]

Which we can use the quadratic formula to solve:

\[ \left( -(-2) \pm \sqrt{(-2) ^{2}-4(1)(2)} \right)\div(2(1))\]

\[\left( -2 \pm \sqrt{-4} \right)\div(2)\]

\[\left( -2 \pm 2i \right)\div(2)\]

\[\left( -1 \pm i \right)\]

So we have (n-(-1-i))(n-(-1+i)

Which simplifies to (n+1+i)(n+1-i).

Now subbing back in the w^2:

\[\left( w ^{2}+1+i \right)\left( w ^{2}+1-i \right)\]

Answer 7

Now, if you want to go a step further, you can use the quadratic formula again:

\[\left( -1\pm \sqrt{0^{2}-4(1)(1+i)} \right)\div(2(1))\]

\[\left( -1\pm \sqrt{-4(1+i)} \right)\div(2)\]

\[\left( -1\pm 2i \sqrt{(1+i)} \right)\div(2)\]

\[ -1/2\pm i \sqrt{(1+i)} \]

And similarly, your other solution will be

\[ -1/2\pm i \sqrt{(1-i)} \]

So your final solution is:

\[ -1/2\pm i \sqrt{(1\pm i)} \]