Question

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

Answer 1

The equation in option A is correct.

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=√(f\left(a\right)),\:\:-√(f\left(a\right))`

`x=√(-1),\:x=-√(-1)`

`x=i,\:x=-i` ∵
`\:√(-1)=i`

solving

`x^2=-12`

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

`x=√(-12),\:x=-√(-12)`

`x=2√(3)i,\:x=-2√(3)i`

Hence, the solutions are:

`x=i,\:x=-i,\:x=2√(3)i,\:x=-2√(3)i`

Therefore, the equation in option A is correct.