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