Question

PLEASE HELP ME!

Answer 1

Answer:

Real roots:

x = {-5, 5}

Imaginary roots:

x = {-4i, 4i}

Step-by-step explanation:

We can solve this via factoring:

$-3x^4+27x^2+1200=0\\\\-3(x^4-9x^2-400)=0\\\\-3(x^4+16x^2-25x^2-400)=0\\\\-3(x^2(x^2+16)-25(x^2+16))=0\\\\-3(x^2-25)(x^2+16)=0\\\\-3(x^2-(5)^2)(x^2+16)=0)\\\\-3(x-5)(x+5)(x^2+16)=0$

And so by solving for x we have:

Case 1:

$x-5=0\\\\x=5$

Case 2:

$x+5=0\\\\x=-5$

Case 3:

$x^2+16=0\\\\x^2=-16\\\\x=\sqrt{-16} \\\\x=\pm4i$

And so you have two real solutions and two imaginary solutions:

Real: $x=\pm 5$

Imaginary: $x=\pm 4i$