# Quadratic with complex roots

• January 10th 2010, 01:51 PM
andyw
Quadratic with complex roots
Hi there,

I'm looking for some confirmation that i have done this correctly/the easiest way,

Rgds.

Resolve $9x^2-36x+37$ into factors.

= 9( $x-\alpha$)( $x-\beta$)

$\frac{36+or-\sqrt{36^2-4*9*37}}{18}$

$\frac{36+or-\sqrt{-36}}{18}$

$\frac{36+or-j6}{18}$

$2+or-j\frac{1}{3}$

$9(x-2+j\frac{1}{3})(x-2-j\frac{1}{3})$

$3(x-2+j\frac{1}{3})3(x-2-j\frac{1}{3})$

$(3x-6+j)(3x-6-j)$
• January 10th 2010, 01:57 PM
pickslides
Quote:

Originally Posted by andyw
Hi there,

I'm looking for some confirmation that i have done this correctly/the easiest way,

to confirm it is correct simply expand your answer

$(3x-6+j)(3x-6-j)$

does it give?

$9x^2-36x+37$
• January 10th 2010, 03:06 PM
andyw
It certainly does!

Thanks.