Find $m$ such that polynomial has real roots

Question

We have the following equation: $$2x^3+3mx^2-m=0$$ Find $m\in\mathbb{R}$ such that the above equation has three real solutions.

My first attempt was to find the first derivative and set the discriminant greater or equal to zero. Doing this I get that $m$ can be any real number. Is this ok?

Answer 1

HINT

Let consider

• $f'(x)=6x^2+6mx=0\implies x=0 \quad x=-m$

then consider $m$ values such that

• $f(-m)\cdot f(0)<0$

Answer 2

Hint: write $$m=-\frac{2x^3}{3x^2-1}$$ and define $$f(x)=-\frac{2x^3}{3x^2-1}$$ $$f'(x)=-6\,{\frac {{x}^{2} \left( -1+x \right) \left( x+1 \right) }{ \left( 3\,{x}^{2}-1 \right) ^{2}}} $$

Answer 3

The cubic $f(x)=2x^3+3mx^2-m$ has always an extrema for $x=0$ because $f'(x)=6x^2+6mx$. Besides $(x,m)=(-1,1)$ implies $f(x)=f'(x)=0$ and the cubic has a maximum at $x=1$ and a minimum at $x=0$. Similarly with $(x,m)=(1,-1)$.

For the shape of the cubic $f(x)$, there are three real roots when $m\ge 1$ and $m\le 1$ or equivalently $m\notin (-1,1)$ (open interval).