Answer:
The zeros of the equation [tex]-3x^4 + 27x^2 + 1200 = 0[/tex] are 4i,-4i,5,-5.
Step-by-step explanation:
Given : Equation [tex]-3x^4 + 27x^2 + 1200 = 0[/tex]
To find : All the zeroes of the equation ?
Solution :
[tex]-3x^4 + 27x^2 + 1200 = 0[/tex]
First we divide whole equation by 3,
[tex]-x^4 +9x^2 +400 = 0[/tex]
Now, Let [tex]x^2=y[/tex]
So, [tex]-y^2 +9y+400 = 0[/tex]
Solve by quadratic formula, [tex]y=\frac{-b\pm\sqrt{b^2-4ac}}{2a}[/tex]
Here, a=-1, b=9 and c=400
[tex]y=\frac{-9\pm\sqrt{9^2-4(-1)(400)}}{2(-1)}[/tex]
[tex]y=\frac{-9\pm\sqrt{1681}}{-2}[/tex]
[tex]y=\frac{-9\pm 41}{-2}[/tex]
[tex]y=\frac{-9+41}{-2},\frac{-9-41}{-2}[/tex]
[tex]y=-16,25[/tex]
Substituting back,
[tex]x=\pm\sqrt{y}[/tex]
When y=-16,
[tex]x=\pm\sqrt{-16}[/tex]
[tex]x=\pm 4i[/tex]
When y=25,
[tex]x=\pm\sqrt{25}[/tex]
[tex]x=\pm 5[/tex]
The zeros of the equation [tex]-3x^4 + 27x^2 + 1200 = 0[/tex] are 4i,-4i,5,-5.
