I am asked to solve for x in the polynomial using factoring and grouping:


My working:




So: $X^2+9=0$



The other solution is $5/2$

My question is, I arrive at just $3i$ whereas my textbook solution says it's $\pm3i$.

How could $-3i$ be a solution here when the input is $i\sqrt{9}$?

  • 2
    $\begingroup$ Because $\sqrt{-9}=\pm3i$. In particular, $(3i)^2=-9$ and $(-3i)^2=-9$. $\endgroup$ – Clayton May 22 '19 at 23:18
  • 1
    $\begingroup$ Whenever you take the square root over both sides of an equality like $x^2 = a$, you always get two solutions: $x = \pm \sqrt{a}$. This is because a negative squared is always a positive, so you always get a negative root as well as the principal root! $\endgroup$ – Jack Crawford May 22 '19 at 23:19
  • $\begingroup$ That makes sense. I'm used to thinking in terms of the root of every regular number has both positive and negative solutions, never considered it works both ways and using i doesn't change that $\endgroup$ – Doug Fir May 22 '19 at 23:30
  • $\begingroup$ So $\pm3i$ are both solutions, though you can’t say one is positive and one is negative $\endgroup$ – J. W. Tanner May 22 '19 at 23:32
  • 1
    $\begingroup$ By the way, the zero of $5x-2$ is $\frac25$, not $\frac52$ as you wrote $\endgroup$ – J. W. Tanner May 22 '19 at 23:42

In complex numbers, there are two zeroes of $x^2-c,$ except only $x=0$ when $c=0.$ In particular, as indicated in comments, the zeroes of $x^2+9$ are $\pm3i.$

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Because $$(-t)^2\equiv(-1)^2(t)^2\equiv t^2$$ hence $i^2\equiv(-i)^2$ and so solutions to quadratics must have both the positive and negative.

Let's prove this, suppose $z=x+iy$ solves $f(z)=0$, where $f(z)=az^2+bz+c; a, b, c\in\Bbb R$

$$z=x+iy \to z^2 =(x^2-y^2) + (2xy)i$$

$$\to ax^2-ay^2 + 2axyi +bx +byi +c =0$$ $$\implies ax^2-ay^2+bx+c=0 \text{ and } 2axy+by=0$$

For $\bar{z}=x-iy$, we have $\bar{z}^2=(x^2-y^2)-2xyi$ and $$f(\bar{z})=ax^2-ay^2-2axyi +bx-byi+c$$ $$=(ax^2-ay^2+bx+c)+i(-2axy-by)$$

We clearly see that $\Re(f(\bar{z}))=\Re(f(z))=0$ and $\Im(f(\bar{z}))=-\Im(f(z))=0$

Hence $f(z)=0 \implies f(\bar{z})=0$

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  • $\begingroup$ You mean when $a,b,c\in\Bbb R$? $\endgroup$ – J. W. Tanner May 23 '19 at 0:16
  • $\begingroup$ Yes, I see that's a requirement for this proof. $\endgroup$ – Rhys Hughes May 23 '19 at 5:41

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