Answer:
The solutions are:
x=0, x=-2, x=4i, x=-4i
Step-by-step explanation:
We need to find solutions of the equation [tex]- 5x^4 + 16x^2 + 32x= - 6x^4 - 2x^3[/tex]
Note in given question: [tex]- 5x^4 + 16x? + 32x= - 6x^4 - 2x^3[/tex] considering 2 instead of ?
Solving:
[tex]- 5x^4 + 16x^2 + 32x= - 6x^4 - 2x^3[/tex]
[tex]- 5x^4 + 16x^2 + 32x+ 6x^4 + 2x^3=0\\- 5x^4+ 6x^4 + 16x^2 + 32x + 2x^3=0\\x^4+2x^3+16x^2+32x=0[/tex]Taking x common
[tex]x(x^3+2x^2+16x+32)=0\\x=0 \ or \ x^3+2x^2+16x+32=0[/tex]
[tex]Simplifying \ x^3+2x^2+16x+32=0 \ we \ get (x+2)(x^2+16)[/tex]
[tex]x=0 \ or \ (x+2)(x^2+16)=0\\x=0 \ or \ x+2=0 \ or x^2+16=0\\x=0 \ or \ x=-2 \ or x^2=-16\\We \ know \ \sqrt{-1}=i \\x=0 \ or \ x=-2 \ or x=\pm 4i\\[/tex]
So, the solutions are:
x=0, x=-2, x=4i, x=-4i
