Blast from the past: remember the quadratic formula

\[

x = \frac{-b \pm \sqrt{{b}^{2}-4ac}}{2a}

\]

and you would look at the discriminant (the *b*² – 4*ac* part) to determine if there were 1, 2, or no solutions? You were told that if *b*² – 4*ac* < 0, then there were no solutions. Well that was a lie (please don’t ask me about Santa Claus). It turns out that there are always solutions to a quadratic equation

*ax*² + *bx* + *c* = 0, where *a*, *b*, and *c* are real and *a* ≠ 0.

Let’s look at *z*² + 4*z* + 29 = 0 and solve for *z*. Using the quadratic formula, we get

\[z = \frac{-4 \pm \sqrt{{4}^{2} – 4(1)(29)}}{2(1)} = \frac{-4 \pm \sqrt{-100 }}{2}\]

The square root of -100 used to be a problem but no longer:

\[\frac{-4 \pm \sqrt{-100 }}{2} = \frac{-4 \pm \sqrt{100 }\sqrt{-1}}{2} = = \frac{-4 \pm 10i}{2}= -2 \pm 5i \]

Notice that the answer is a complex conjugate pair. This always happens for any polynomial equation with real coefficients. If a complex number is a root, so is its conjugate. The answer above also means that

*z*² + 4*z* + 29 = (*z* + 2 – 5*i*)(*z* + 2 + 5*i*)

So now if *b*² – 4*ac* < 0, the quadratic equation has a conjugate pair as the solution.

If solutions to polynomial equations are complex conjugates, that means that a cubic equation has to have at least one real root because a cubic equation can have at most, three solutions. In fact, any odd powered polynomial equation has to have at least one real root. This makes sense graphically because the highest power of an odd-powered polynomial will eventually dominate all the other terms as *x* goes very negative or very positive. The highest powered term will have opposite signs for negative and positive values of *x*, so the polynomial will eventually cross the *x*-axis.

Let’s do one more problem. If a polynomial has single roots 2 ± *i* and 3, what is that polynomial? To solve this, just form the factors with these roots:

(*z* -2 –*i*)(*z* -2 +*i*)(*z* -3) = (*z*² – 4*z* + 5)(*z* -3) = *z*³ – 7*z*² + 17*z* – 15

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