Cassarrim1

2022-01-21

How do you solve ${z}^{2}+12=-z$?

spelkw

Expert

${z}^{2}+12=-z⇔{x}^{2}+x+12=0$
$z=\frac{-1±\sqrt{{1}^{2}-4\left(12\right)\left(1\right)}}{2}=-\frac{1}{2}±\frac{\sqrt{47}}{2}i$
Alternatively, you could complete the square,
${z}^{2}+z+12=0$
${z}^{2}+z+\frac{1}{4}+\frac{47}{4}=0$
${\left(z+\frac{1}{2}\right)}^{2}=-\frac{47}{4}$
$z+\frac{1}{2}=±\frac{\sqrt{47}}{2}i$
$z+\frac{1}{2}=±\frac{\sqrt{47}}{2}i$
$z=-\frac{1}{2}±\frac{\sqrt{47}}{2}i$
Which yields the same result (the quadratic formula is derived from completing the square)

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