Factorise

Posted July 26th, 2008 by alpha

$\sqrt{3}x^2-2\sqrt{2}x-2\sqrt{3}=0$

Comments

On July 28th, 2008 Structure says:

$\sqrt{3}x^2-2\sqrt{2}x-2\sqrt{3}=\sqrt3 x^2+\sqrt2 x-3\sqrt2x-2\sqrt3=x(\sqrt3 x+\sqrt2 )-\sqrt6(\sqrt3 x+\sqrt2 )=$

$=(\sqrt3 x+\sqrt2 )(x-\sqrt 6)=0$

$x_1=-\frac{\sqrt 2}{\sqrt 3}=-\frac{\sqrt 6}{3}$

$x_2=\sqrt 6$

Depending on your interest you may apply the formula of solving secorn order algebraic equation.

$ax^2+bx+c=a(x-x_1)(x-x_2)$

So you can write

$x_{1,2}=\frac{-b\pm\sqrt{b^2-4ac}}{2a}$

In your example

$x_{1,2}=\frac{2\sqrt 2\pm\sqrt{8+24}}{2\sqrt 3}=\frac{2\sqrt 2\pm4\sqrt{2}}{2\sqrt 3}=\frac{\sqrt 2\pm2\sqrt{2}}{\sqrt 3}$