Finding complex roots of polynomials

**Rae** · November 28th 2010, 01:49 PM

Find all solutions of the equation:

x^4 + 2x^2 - 8 = 0

I know the roots are either complex or irrational because none of the rational possibilities work out. Is there a way to find the other roots without using a graphing calculator? Thanks for any help.

**pickslides** · November 28th 2010, 01:58 PM

start by saying $a = x^2$ then solve $a^2 + 2a - 8 = 0$

what do you get?

**e^(i*pi)** · November 28th 2010, 01:59 PM

Let $u = x^2$

$u^2 + 2u - 8 = (u-2)(u+4) = 0$

$u = x^2 = 2 \: \implies x = \pm \sqrt{2}$

OR

$u = x^2 = -4 \: \implies x = \pm 2i \: \text{ where } i^2 = -1$

**dwsmith** · November 28th 2010, 02:02 PM

Also, by Descartes Rule of Signs, we know there are 1 positive and 1 negative solution; hence, we have a pair of complex numbers.

If you use the formula $\displaystyle \frac{P}{Q}$ , you can use synthetic division to find your 2 real solutions and then use $\displaystyle x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}$ to find the complex pair.

**Rae** · November 28th 2010, 02:22 PM

Originally Posted by dwsmith

Also, by Descartes Rule of Signs, we know there are 1 positive and 1 negative solution; hence, we have a pair of complex numbers.

If you use the formula $\displaystyle \frac{P}{Q}$ , you can use synthetic division to find your 2 real solutions and then use $\displaystyle x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}$ to find the complex pair.

Doesn't the formula P/Q only find the rational real roots and not the irrational?

**dwsmith** · November 28th 2010, 02:24 PM

You equation doesn't contain irrational values.

**Rae** · November 28th 2010, 02:28 PM

The equation has 2 complex roots and 2 irrational roots. I tried using P/Q and none of the rational possibilties were roots.

**Plato** · November 28th 2010, 02:30 PM

Why not tackle it head on?

Factor it: $(x^2+4)(x^2-2)$ .

From that the roots pop right out: $x=\pm 2i~\&~\pm\sqrt2~.$

**dwsmith** · November 28th 2010, 02:31 PM

Originally Posted by Rae

The equation has 2 complex roots and 2 irrational roots. I tried using P/Q and none of the rational possibilties were roots.

You are correct. I really didn't look at the question. haha

**Rae** · November 28th 2010, 02:33 PM

Thanks, I understand how to find the solutions now.

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