Question

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

Answer 1

`\bf -3x^4+27x^2+1200=0\implies -3(x^4-9x^2-400)=0 \\\\\\ x^4-9x^2-400=0\implies (x^2-25)(x^2+16)=0\implies \stackrel{\textit{difference of squares}}{\stackrel{\downarrow }{(x^2-5^2)}(x^2+16)=0} \\\\\\ (x-5)(x+5)(x^2+16)=0\implies \begin{cases} x-5=0\implies &x=5\\ x+5=0\implies &x=-5 \end{cases}`

now, as far as x² + 16 = 0 root, well

`\bf x^2+16=0\implies x^2=-16\implies x=\pm√(-16)\implies x=\pm 4i`

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

Answer 2

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

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