Answer-
The zeros are, [tex]5,\ -5,\ 4i,\ -4i[/tex]
Solution-
[tex]\Rightarrow -3x^4+27x^2+1200=0[/tex]
[tex]\Rightarrow -3(x^2)^2+27(x^2)+1200=0[/tex]
Here,
a = -3, b = 27, c = 1200
So,
[tex]x^2=\dfrac{-b\pm \sqrt{b^2-4ac}}{2a}[/tex]
[tex]=\dfrac{-27\pm \sqrt{-27^2-4\cdot (-3)\cdot 1200}}{2\cdot (-3)}[/tex]
[tex]=\dfrac{-27\pm \sqrt{729+14400}}{-6}[/tex]
[tex]=\dfrac{-27\pm 123}{-6}[/tex]
[tex]=\dfrac{-27+123}{-6},\ \dfrac{-27- 123}{-6}[/tex]
[tex]=\dfrac{96}{-6},\ \dfrac{-150}{-6}[/tex]
[tex]=-16,\ 25[/tex]
So,
[tex]\Rightarrow x^2=25,\ -16[/tex]
[tex]\Rightarrow x=\sqrt{25},\ \sqrt{-16}[/tex]
[tex]\Rightarrow x=5,\ -5,\ 4i,\ -4i[/tex]
