Find all zeros of the equation -3x^4+27x^2+1200=0 step by step

The roots are +/- 5 and +/- 4i  or 5, -5,  4i,  -4i

Step-by-step explanation:

Begin by dividing through by - 3

-3x^4 / - 3   + 27x^2/-3   +   1200/-3 = 0

x^4 - 9x^2 - 400 = 0                   This makes it easier to work with. Substitute

z = x^2

z^2 - 9z - 400 = 0                        Factor

(z - 25)(z  + 16) =  0                       Put the x^2 back in

(x^2 - 25)(x^2 + 16)

Factor x^2 - 25 using the difference of squares.

(x^2 - 25) = (x + 5)(x - 5)

x = +5 and x = - 5

Factor x^2 + 16

x^2 + 16 = 0

x^2 = - 16                Take the square root of x^2 and -16

x = sqrt(-16)

x = +4i, x = - 4i

Answer

The roots are +/- 5 and +/- 4i

now, as far as x² + 16 = 0 root, well

which is another way of saying, there's no solution coming from that factor, or one can also say, there are two complex solutions.

x = 2

step-by-step explanation:

if sr parallel to ba then the sides bc and ac are in proportion, that is

= ← substitute values

= ( cross- multiply )

27x = 6(4x + 1)

27x = 24x + 6 ( subtract 24x from both sides )

3x = 6 ( divide both sides by 3 )

x = 2