# Equation

• Nov 18th 2007, 10:27 AM
fluffy_penguin
Equation
http://i3.tinypic.com/86q0wgh.jpg

$\displaystyle 9x^2-12x+1=0$

And since I couldn't go any further with it I thought it might be no solution

$\displaystyle (3x -?)(3x -?)$
• Nov 18th 2007, 10:34 AM
Jhevon
Quote:

Originally Posted by fluffy_penguin
http://i3.tinypic.com/86q0wgh.jpg

$\displaystyle 9x^2-12x+1=0$

And since I couldn't go any further with it I thought it might be no solution

$\displaystyle (3x -?)(3x -?)$

you're doing complex numbers now right? which means your quadratics will always have solutions, just perhaps, not real ones.

if factoring is a problem, use the quadratic formula
• Nov 18th 2007, 10:42 AM
Krizalid
$\displaystyle 9x^2-12x+1=(3x-2)^2-3.$
• Nov 18th 2007, 10:54 AM
fluffy_penguin
To be honest, I'm still confused. We're on the radical chapters, I don't know why I got asked that question. But I'll try the quadratic forumla for it.
• Nov 18th 2007, 10:57 AM
Krizalid
Sure.

Then you want an expression like this

$\displaystyle (3x - 2)^2 - 3 = (3x - 2)^2 - \left( {\sqrt 3 } \right)^2 = \left( {3x - 2 + \sqrt 3 } \right)\left( {3x - 2 - \sqrt 3 } \right).$