Solving a polynomial by grouping and factoring - why does this answer have $\pm3i$?

Question

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

$5X^3+45X=2X^2+18$

My working:

$5X^3-2X^2+45X-18$

$X^2(5X-2)+9(5X-2)$

$(X^2+9)(5X-2)$

So: $X^2+9=0$

$X^2=-9$

$X=i\sqrt{9}=3i$

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 $i\sqrt{9}$?

Answer 1

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 $\pm3i.$

Answer 2

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