Question

Which equation has x = -i and x = = 21v3 as two of its solutions?

Answer 1

Answer:

The equation in option A is correct.

Step-by-step explanation:

Given the equation

$\:y=x^4+13x^2+12$

Let us find the solution by setting y=0

$x^4+13x^2+12=0\:\:\:\:\:$

$\mathrm{Rewrite\:the\:equation\:with\:}u=x^2\mathrm{\:and\:}u^2=x^4$

$u^2+13u+12=0$

if $ab=0\:\mathrm{then}\:a=0\:\mathrm{or}\:b=0\:\left(\mathrm{or\:both}\:a=0\:\mathrm{and}\:b=0\right)$

$u+1=0\quad \mathrm{or}\quad \:u+12=0$

$u=-1,\:u=-12$

$\mathrm{Substitute\:back}\:u=x^2,\:\mathrm{solve\:for}\:x$

$x^2=-1,\:x^2=-12$

solving

$x^2=-1$

$\mathrm{For\:}x^2=f\left(a\right)\mathrm{\:the\:solutions\:are\:}x=\sqrt{f\left(a\right)},\:\:-\sqrt{f\left(a\right)}$

$x=\sqrt{-1},\:x=-\sqrt{-1}$

$x=i,\:x=-i$            ∵ $\:\sqrt{-1}=i$

solving

$x^2=-12$

$\mathrm{For\:}x^2=f\left(a\right)\mathrm{\:the\:solutions\:are\:}x=\sqrt{f\left(a\right)},\:\:-\sqrt{f\left(a\right)}$

$x=\sqrt{-12},\:x=-\sqrt{-12}$

$x=2\sqrt{3}i,\:x=-2\sqrt{3}i$

Hence, the solutions are:

$x=i,\:x=-i,\:x=2\sqrt{3}i,\:x=-2\sqrt{3}i$

Therefore, the equation in option A is correct.