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

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 $$i\sqrt{9}$$?

• Because $\sqrt{-9}=\pm3i$. In particular, $(3i)^2=-9$ and $(-3i)^2=-9$. – Clayton May 22 '19 at 23:18
• 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! – Jack Crawford May 22 '19 at 23:19
• 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 – Doug Fir May 22 '19 at 23:30
• So $\pm3i$ are both solutions, though you can’t say one is positive and one is negative – J. W. Tanner May 22 '19 at 23:32
• By the way, the zero of $5x-2$ is $\frac25$, not $\frac52$ as you wrote – 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 $$\pm3i.$$

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

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