Answer:
The equation in option A is correct.
Step-by-step explanation:
Given the equation
[tex]\:y=x^4+13x^2+12[/tex]
Let us find the solution by setting y=0
[tex]x^4+13x^2+12=0\:\:\:\:\:[/tex]
[tex]\mathrm{Rewrite\:the\:equation\:with\:}u=x^2\mathrm{\:and\:}u^2=x^4[/tex]
[tex]u^2+13u+12=0[/tex]
if [tex]ab=0\:\mathrm{then}\:a=0\:\mathrm{or}\:b=0\:\left(\mathrm{or\:both}\:a=0\:\mathrm{and}\:b=0\right)[/tex]
[tex]u+1=0\quad \mathrm{or}\quad \:u+12=0[/tex]
[tex]u=-1,\:u=-12[/tex]
[tex]\mathrm{Substitute\:back}\:u=x^2,\:\mathrm{solve\:for}\:x[/tex]
[tex]x^2=-1,\:x^2=-12[/tex]
solving
[tex]x^2=-1[/tex]
[tex]\mathrm{For\:}x^2=f\left(a\right)\mathrm{\:the\:solutions\:are\:}x=\sqrt{f\left(a\right)},\:\:-\sqrt{f\left(a\right)}[/tex]
[tex]x=\sqrt{-1},\:x=-\sqrt{-1}[/tex]
[tex]x=i,\:x=-i[/tex] ∵ [tex]\:\sqrt{-1}=i[/tex]
solving
[tex]x^2=-12[/tex]
[tex]\mathrm{For\:}x^2=f\left(a\right)\mathrm{\:the\:solutions\:are\:}x=\sqrt{f\left(a\right)},\:\:-\sqrt{f\left(a\right)}[/tex]
[tex]x=\sqrt{-12},\:x=-\sqrt{-12}[/tex]
[tex]x=2\sqrt{3}i,\:x=-2\sqrt{3}i[/tex]
Hence, the solutions are:
[tex]x=i,\:x=-i,\:x=2\sqrt{3}i,\:x=-2\sqrt{3}i[/tex]
Therefore, the equation in option A is correct.
