# quadratics and imaginary roots

• Aug 13th 2012, 09:03 AM
anthonye
quadratics and imaginary roots
Hi;
If a quadratic polynomial won't factor does this mean it
has imaginary roots?

if I complete the square on the same polynomial which makes it
factorable does this process give it real roots?
• Aug 13th 2012, 09:15 AM
Plato
Re: quadratics and imaginary roots
Quote:

Originally Posted by anthonye
Hi;
If a quadratic polynomial won't factor does this mean it
has imaginary roots?
if I complete the square on the same polynomial which makes it
factorable does this process give it real roots?

Consider $\displaystyle 3x^2-3x-2$, is that what you mean?
• Aug 13th 2012, 09:34 AM
anthonye
Re: quadratics and imaginary roots
Yes
• Aug 13th 2012, 09:41 AM
Prove It
Re: quadratics and imaginary roots
Quote:

Originally Posted by anthonye
Hi;
If a quadratic polynomial won't factor does this mean it
has imaginary roots?

If it won't factor over the reals, then yes, it will have COMPLEX roots (which have a real part and an imaginary part).

Quote:

if I complete the square on the same polynomial which makes it
factorable does this process give it real roots?
Completing the square won't magically make a quadratic factorise over the reals. You WILL however be able to factorise over the complex numbers, which will still give complex roots.
• Aug 13th 2012, 09:44 AM
anthonye
Re: quadratics and imaginary roots
Ok thank you Plato and prove it.
• Aug 13th 2012, 10:24 AM
Plato
Re: quadratics and imaginary roots
Quote:

Originally Posted by anthonye
Yes

The point is that any quadratic with real coefficients can be factored.
Take the example I gave you. $\displaystyle 3x^2-3x-2=0$ has two real roots: $\displaystyle \left( {\frac{{3 + \sqrt {33} }}{6}} \right)\;\& \;\left( {\frac{{3 - \sqrt {33} }}{6}} \right)$.

The factored form is: $\displaystyle 3x^2-3x-2=\left[ {x - \left( {\frac{{3 + \sqrt {33} }}{6}} \right)} \right]\left[ {x - \left( {\frac{{3 - \sqrt {33} }}{6}} \right)} \right]=0$.
• Aug 13th 2012, 11:53 AM
HallsofIvy
Re: quadratics and imaginary roots
On the other hand, $\displaystyle x^2+1$ has no real zeros but also can be factored: (x- i)(x+ i). The trouble is, you didn't say what you meant by "factored".
• Aug 15th 2012, 05:00 AM
anthonye
Re: quadratics and imaginary roots
Please can you elaborate on the fact that any quadratic with real coefficients can be factored

I'm a little lost with that.
• Aug 15th 2012, 05:28 AM
HallsofIvy
Re: quadratics and imaginary roots
Any quadratic, whether its coefficients are real or not, can be factored- but NOT necessarily with integer or even real coefficients.

Take $\displaystyle ax^2+ bx+ c$ as a generic example. First, factor out the "a". $\displaystyle a(x^2+ \frac{b}{a}x+ \frac{c}{a})$. Now "complete the square"- add and subtract $\displaystyle \frac{b^2}{4a^2}$: $\displaystyle a(x^2+ \frac{b}{a}x+ \frac{b^2}{4a}+ \frac{c}{a}- \frac{b^2}{4a^2})$. The point of that is that $\displaystyle x^2+ \frac{b}{a}x+ \frac{b^2}{4a^2}= (x+ \frac{b}{2a})^2$, a perfect square. And of course, $\displaystyle \frac{b^2}{4ac^2}-\frac{c}{a}= \left(\sqrt{\frac{b^2}{4a^2}- \frac{c}{a}}\right)^2$.

That is, we can write $\displaystyle ax^2+ bx+ c= a\left(\left(x+ \frac{b}{2a}\right)^2- \left(\sqrt{\frac{b^2}{4a^2}- \frac{c}{a}}\right)^2\right)$

And now we use the fact that $\displaystyle x^2- a^2= (x- a)(x+ a)$ to write
$\displaystyle a\left(x+ \frac{b}{2a}+\sqrt{\frac{b^2}{4a^2}- \frac{c}{a}}\right)\left(x+ \frac{b}{2a}- \sqrt{\frac{b^2}{4a^2}- \frac{c}{a}}\right)$

That is factored into a number times two linear factors. With those square roots, those numbers may not be integers or even real numbers.
• Aug 15th 2012, 05:32 AM
anthonye
Re: quadratics and imaginary roots
Wow ok thanks Hallsofivy Ineed to look at this more any idea where I can read about this subject?