Find the solution of this equation: x^3 + 6x = 3*sqrt(3) * x^2

Question

directrix · Answer

What are the instructions for this problem?

directrix · Answer

If you are asked to find the roots, see if the expression will factor over the set of irrational numbers.

zyberg · Answer

Yes, @Directrix , I am asked to find the roots.

eliesaab · Answer

I will give you a hint to start doing it.
Is x=0 a root?

eliesaab · Answer

You can write it as
$$
x^3+ 3 x= 3 \sqrt 3 x^2\
x^3+ 3 x-3 \sqrt 3 x^2=0\
x( x^2 +3 -3 \sqrt 3 x)=0

$$
You can finish it now

eliesaab · Answer

Use the quadratic formula for the quadratic

directrix · Answer

@eliesaab  Do you see my error?

When I factored out the x, I got:

x^3 + 6x = 3*sqrt(3) * x^2 =

x*( x² - ( 3√3 ) *x + 6 )

-------

( x² - ( 3√3 ) *x + 6 ) will factor over the set of irrationals.

eliesaab · Answer

I copied the 6x as 3 x, my mistake

$$
x^3+ 6 x= 3 \sqrt 3 x^2\
x^3+ 6 x-3 \sqrt 3 x^2=0\
x( x^2 +6 -3 \sqrt 3 x)=0
$$

eliesaab · Answer

$$
x^2-3 \sqrt{3} x+6=\left(x-2 \sqrt{3}\right) \left(x-\sqrt{3}\right)
$$