I'm going through the AoPS Algebra book, and I'm on the quadratics section.

I'm given this challenge question:

$ \displaystyle 2x^2 + 7x(\sqrt{3}) + 9 = 0$

And I have to solve for all values of $x$.

How I try to do it:

Simplify the notation of the numbers by converting each number to radical notation, and then solve by factoring the quadratic as would be done normally, except use the simpler notation of the numbers to your advantage. This seems all right, but every time I try to go from there on, I encounter a difficulty as I am no longer able to factor despite me having only converted numbers to radical form... Could someone help me on how to approach this question? I know it's easy but I'm just trying to improve my skills.



${\, 0\, =\, 2x^2\!+7\sqrt{3}x + 9 \underset{\large x\, =\, \sqrt{3}\,y}{=}\! 3(2y^2\!+7y+3) \,=\, 3(y+3)(2y+1)}\,$ by the AC-method.

Therefore $\ y = -3;\ \ \ {-}1/2\ $
so $\ x = \sqrt{3}y = -3\sqrt{3};\, -\sqrt{3}/2$

Remark $ $ We could also rotely apply the quadratic formula, but generally the above method will be easier since it eliminates irrationals, and reduces the size of the numbers in calculations.

  • $\begingroup$ Hmm, don't really understand how you did something. I'm not sure how you go from $2x^2 + 7(sqrt3)x + 9$ to $3(2y^2 + 7y + 3)$, how'd the 9 go from 9 to 3? You're multiplying by 3, wouldn't you get 27? $\endgroup$ – user164403 Aug 28 '14 at 21:34
  • $\begingroup$ @user164403 Please tell me what is not clear and I'll be happy to elaborate. $\endgroup$ – Bill Dubuque Aug 28 '14 at 21:35
  • $\begingroup$ Edited the above comment. $\endgroup$ – user164403 Aug 28 '14 at 21:39
  • $\begingroup$ @user164403 To get rid of the $\sqrt3$ we change variables $\, x = \sqrt{3}y.\,$ Do that substitution. Then $\,x^2 = 3y^2,\ \sqrt{3} x = 3y,\ \ldots\ $ then pull out a factor of $\, 3\ $ from the resulting polynomial in $\,y.\ $ $\endgroup$ – Bill Dubuque Aug 28 '14 at 21:41
  • $\begingroup$ I see it!! Thanks $\endgroup$ – user164403 Aug 28 '14 at 21:48


$$ (2x + \sqrt 3)(x + 3\sqrt 3) $$

Note that when you have a prime $2$ as the coefficient of $x^2$ the factors are almost always of the form $(2x \pm ..)(x \pm ..)$ because $2$ and $1$ are the only factors of $2$(unless of course there are more radicals present). And the $\sqrt 3$ should be incorporated into $9$. That is how you work it.

  • $\begingroup$ @user164403 One can eliminate guesswork by using the AC-method (algorithm) - follow the link in my answer. This is essential for less trivial problems. $\endgroup$ – Bill Dubuque Aug 24 '14 at 16:28

Hopefully this! Methid will help you again


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