Solving A Cubic Equation With An Irrational Root

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[Nepenth]

Substitute x=sqrt(2)*z and you get the cubic equation sqrt(2) * (2z^3 + 3z + 5) = 0 with integer coefficients. Obviously z=-1 is a solution (rational root theorem!) and dividing (2z^3 +3z +5) by ((z+1) leaves us with a quadratic equation in z, which can easily be solved.

[andreabaldacci]

The function f(X)=X³+3X is always increasing. This means that -5sqrt(2) is met at only one point for X<0. Then I solved X³+3X+5sqrt(2)=(X-A)(X²+BX+C), or equivalently the system B=A, C-AB=3, -AC=sqrt(2). This leads to C=5, A=-sqrt(2), B=sqrt(2). Hence X=-sqrt(2), and from X²-sqrt(2)X+5=0 X=two extra complex solutions.

[allanmarder]

x^3 +3x + 5sqrt(2) =0 is a depressed cubic of the form x^3 + p*x +q where p=3 and q=5*sqrt(2). The discriminant D = (q/2)^2 + (p/3)^3 is >0.
That tells us there is only 1 real root. The depressed cubic formula is x = (-q/2 + sqrt(D))^(1/3) + (-q/2 - sqrt(D))^(1/3). I used a calculator to compute
this and the answer to 3 decimals was -1.414 Looking at this I assumed (as did Syber) the root was related to the sqrt(2). I put x= -sqrt(2) in the formula
and it checked. I wrote the original equation as (x+sqrt(2))(x^2 -sqrt(2)x +5) =0 The 2 complex roots were then obtained by the quadratic formula.

[dvgvhut]

x^3+3x+5rt2=0
(x+rt2)(x^2-rt2 x+5)=0
x=-rr2.
You can find the two complex solutions solutions from the equation of
x^2-rt2 x+5=0

[rob]

Why not just divide like this in last section?
x^3 + 3x + 5√2 = x^2(x + √2) - x√2(x + √2) + 5(x + √2)
= (x + √2)(x^2 - √2x + 5)