# Minimum value of $a+b$

If the graph of $f(x)=2x^3+ax^2+bx$ intersects the $x$-axis at three distinct points, then what is minimum value of $a+b$? Here $a$ and $b$ are natural numbers.

My attempt:
As the graph intersects the $x$-axis at three distinct points, it has $2$ local maxima/minima. Let these $2$ local maxima/minima be $x_1, x_2$.

I found $f'(x)=6x^2+2ax+b$
So $x_1, x_2$ are the roots of $6x^2+2ax+b=0$

I could not solve this further. I think the minimum value of $a+b$ is $\sqrt {ab}$.

Is my approach correct or is there any other method?

It must cut at $x=0$, so we need to ensure just that $2x^2+ax+b$ has two distinct non-zero roots. Clearly $b \neq 0$.
So the discriminant $\Delta = a^2-8b > 0 \implies a^2 > 8b$. Among natural numbers, it should be obvious we must choose $b=1, a = 3$ to get the minimum $a+b$.
For intersection of the curve $f(x)=2x^3+ax^2+bx$ with the x-axis, we have $$f(x)=0 \implies 2x^3+ax^2+bx=0$$ $$\implies x(2x^2+ax+b)=0 \implies x=0\ \text{&}\ 2x^2+ax+b=0$$ Now, for intersection of curve $f(x)$ with the x-axis at three distinct points, the quadratic equation: $2x^2+ax+b=0$ must have two distinct roots.
Hence, we have determinant $B^2-4AC>0$ $$\implies (a)^2-4(2)(b)>0$$ $$\implies a^2-8b>0$$ $$\implies \color{red}{a^2>8b}$$ Since, $a$ & $b$ are natural numbers hence, to get minimum value of $\color{red}{(a+b)}$ the numbers $a$ & $b$ both must be least natural numbers satisfying the above relation.
Hence, substituting the least possible value of $b$ as $b=1$, we get $$a^2>8\times 1\implies a^2>8$$ Now, the least possible value of $a$ satisfying the above inequality is $8$
Thus, $a=3$ & $b=1$ are fully satisfying the inequality, hence the minimum value of $(a+b)$ is given as follows $$\color{blue}{a+b=3+1=4}$$