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

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

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- Nov 18th 2007, 10:27 AMfluffy_penguinEquation
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 AMJhevon
- Nov 18th 2007, 10:42 AMKrizalid
$\displaystyle 9x^2-12x+1=(3x-2)^2-3.$

- Nov 18th 2007, 10:54 AMfluffy_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 AMKrizalid
Sure.

Then you want an expression like this

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

Now you have radicals there.