Question

Solve for x

3x^4 + 18x^2 + 24 = 0

Answer 1

so far I have
3(x^4 + 6x^2 + 8) = 0

Answer 2

3 is never =0, right? hence, the term inside the bracket =0, right?

Answer 3

oh whoops, you're right

Answer 4

Now, let t = x^2, solve the quadratic in term of t

Answer 5

t^2 + 6t + 8

so

(t + 4)(t + 2)?

Answer 6

I think so

Answer 7

so then really

(x^2 + 4)(x^2 + 2) = 0 and solve for both?

Answer 8

Question: Do you have any field you have to work on? like Real/ Complex?

Answer 9

Both at this point, I'm reviewing my third exam and seeing what I missed, we had reached complex numbers by now

Answer 10

ok, if it is so, you have 4 roots in complex. If not, you don't have any solution in Real

Answer 11

I got \[i \sqrt{2}\] for both

Answer 12

how? for x^2+4 =0, you have 2 roots
for x^2+2 =0 , you have another 2 roots
They are different, right?

Answer 13

Oh wait, I see what I did wrong

Answer 14

x^2 + 4 = 0
-4 -4
x^2 = -4
sqrtx^2 = sqrt-4
x = \[\pm 2i\]

Answer 15

x^2 + 2 = 0
- 2 -2
x^2 = -2
sqrtx^2 = sqrt-2
x = isqrt2?

Answer 16

yup for the first one, no for the second one

Answer 17

\(x=\pm i\sqrt{2}\)

Answer 18

ohh okay. I wasn't sure where to put the plus/minus part, I understand that

Answer 19

Thank you! Would you mind helping me on another one?

Answer 20

let see, if I can