Problems from Definition of complex numbers

Development:

The first one is a complex number, but it is not a pure imaginary since it has a real part. The pure imaginary number is $\sqrt{-27}$. This is because it is a multiple of the imaginary unit.

Solution:

The pure imaginary is $\sqrt{-27}$.

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Development:

If they must have a multiple of the imaginary unit, the easiest thing is to take an imaginary number, equate it to $x$ and then square it: $$12i=x\Rightarrow (12i{)}^2=x^2\Rightarrow 144i^2=x^2\Rightarrow -144=x^2\Rightarrow x^2+144=0$$ it has $12i$ as solution, and $12i$ is a multiple of the imaginary unit $i$. Similarly we can obtain $x^2+169=0$ has a multiple of $i$. The solution would be $13i$.

Solution:

$x^2+144=0$ and $x^2+169=0$.

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Development:

1. $3x^2+27=0\Rightarrow 3x^2=-27\Rightarrow x=\pm \sqrt{-{\displaystyle \frac{27}{3}}}=\pm 3i$
2. $8x^2+4x-2=0\Rightarrow$ ${\displaystyle \begin{array}{rl}x& =\frac{-4\pm \sqrt{16-4\cdot 8\cdot 2}}{16}=\frac{-4\pm \sqrt{-48}}{16}=\frac{-1}{4}\pm \frac{i\sqrt{48}}{16}\\ & =\frac{-1}{4}\pm \frac{i\sqrt{2^4\cdot 3}}{16}=\frac{-1}{4}\pm \frac{i\cdot 2^2\sqrt{3}}{16}=\frac{-1}{4}\pm \frac{i\sqrt{3}}{4}\end{array}}$

Solution:

1. $x=\pm 3i$
2. $x_1={\displaystyle \frac{-1}{4}}+{\displaystyle \frac{\sqrt{3}}{4}}i{x}_2={\displaystyle \frac{-1}{4}}-{\displaystyle \frac{\sqrt{3}}{4}}i$

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