Set $f(x) = ax^3 + bx^2 + cx + d$ We have f'(x)=$3ax^2 + 2bx + c$ f''(x)=$6ax + 2b$ f'''(x)=$6a$ Equation <=>2f(x)f''(x)=(f'(x)$)^2$ Set $g(x)$=2f(x)f''(x)-(f'(x)$)^2$ We have g'(x)=2f(x)f'''(x)=$12a^2(x - \alpha)(x - \beta)(x - \gamma)$ with $\alpha;\beta;\gamma$ is three root of $f(x);\alpha < \beta < \gamma$ Then $g(\alpha)$=-(f'($\alpha$)$)^2 < 0$ $g(\beta)$=-(f'($\beta$)$)^2 < 0$ then this equation has 2 root