Math

Here's another way just by directly factoring....

-3x^4 + 27x^2 + 1200 = 0   multiply through by -1

3x^4 - 27x^2  - 1200  = 0     divide everything by 3

x^4 - 9x^2 - 400  = 0

(x^2  - 25) (x^2 + 16)  = 0      

(x - 5) (x + 5) (x^2 + 16)  =  0        setting the first two factors to 0 we get the two "real" solutions of 5 and - 5

Setting  x^2  +  16  = 0

x^2   = -16        take the square root of both sdes

x  = plus/minus [4 i ]        and these are the two complex [non-real] solutions

Solve for x:
-(3 x^4)+27 x^2+1200 = 0

Substitute y = x^2:
-3 y^2+27 y+1200 = 0

The left hand side factors into a product with three terms:
-3 (y-25) (y+16) = 0

Divide both sides by -3:
(y-25) (y+16) = 0

Split into two equations:
y-25 = 0 or y+16 = 0

Add 25 to both sides:
y = 25 or y+16 = 0

Substitute back for y = x^2:
x^2 = 25 or y+16 = 0

Take the square root of both sides:
x = 5 or x = -5 or y+16 = 0

Subtract 16 from both sides:
x = 5 or x = -5 or y = -16

Substitute back for y = x^2:
x = 5 or x = -5 or x^2 = -16

Take the square root of both sides:
Answer: | x = 5    or x = -5    or x = 4i    or x = -4i
 

Thanks!