Solve w4+2w2-24=0?

Question

Problem:

Solve w(4th power) + 2x(squared) - 24 = 0


Answers:
1) -2, -(sq root)6, (sq root)6,2
2) -2, 2, -i(sq root)6, i(sq root)6
3) -(sq root)6, 2, 2i, i-(sq root)6
4) -2i, 2i, -(sq root)6, (sq root)6

Answer 1

w4+2w2-24=0
w4+(6-4)w2-24=0
w4+6w2-4w2-24=0
w2(w2+6)-4(w2+6)=0
(w2+6)(w2-4)=0
Hence either w2+6=0
or w2-4=0
Hence w = -2, 2, -i(sq root)6, i(sq root)6
Therefore alternate 2 is correct.

Answer 2

2x^2-14x+24=0, divide for the period of with the aid of '2' to make x^2 cohesion. This turns into: x^2-7x+12=0, then we seem for 2 numbers such that as quickly as we multiply them we get +12 and as quickly as we upload them we get -7. The numbers are -3 and -4. So, putting this in determination to -7, we gets; x^2-3x-4x+12=0,=x(x-3)-4(x-3) =0, =(x-3)(x-4)=0, this ability that the two x-3=0 or x-4=0, :. X=3 or 4.

Answer 3

w^4 + 2x^2 - 24 = (w^2 - 4)(w^2 + 6) = (w + 2)(w - 2)(w^2 + 6)

Use the quadratic formula for the last part, and you have

w = 2, -2, i√6, -i√6