# Find $m$ such that polynomial has real roots

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?

• It is not enough to have a derivative with two real roots, as this only means that there are two extrema, which can be of the same sign. – user65203 May 11 '18 at 13:32

HINT

Let consider

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

then consider $m$ values such that

• $f(-m)\cdot f(0)<0$
• That is not the derivative... – asd11 May 11 '18 at 12:47
• @asd11 ops...of course I fix! Thanks – user May 11 '18 at 12:48

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

• Interesting. But I don't follow, what comes next? – asd11 May 11 '18 at 13:00
• You must plot the function $f(x)$ and since $m$ is constant see how often intersect $m$ $f(x)$ – Dr. Sonnhard Graubner May 11 '18 at 13:02
• So if $m>1$ we get three real roots – Dr. Sonnhard Graubner May 11 '18 at 13:05

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).