Question:
$4 x^{2}+1=0$
Solution:
We have:
$4 x^{2}+1=0$
$\Rightarrow(2 x)^{2}-i^{2}=0$
$\Rightarrow(2 x)^{2}-(i)^{2}=0$
$\Rightarrow(2 x+i)(2 x-i)=0$
$\Rightarrow(2 x+i)=0$ or $(2 x-i)=0$
$\Rightarrow 2 x=-i$ or $2 x=i$
$\Rightarrow x=-\frac{i}{2}$ or $x=\frac{i}{2}$
Hence, the roots of the equation are $\frac{1}{2} i$ and $-\frac{1}{2} i$.
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