Is There A Cubic Discriminant?

This is a direct application of cubic discriminant . Yes! There is in fact a cubic discriminant.

If the cubic equation is of the form $ax^3 + bx^2 + cx + d = 0$ , then its discriminant is $D_3 = b^2c^2 - 4ac^3 - 4b^3 d - 27a^2 d^2 + 18abcd$ . Furthermore, if the it has three not necessarily distinct real roots, then $D_3 \geq 0$ .

In this case, we have $a = 2, b = -3, c = 0, d = p$ . Substuting these values into $D_3$ gives

$\begin{aligned} D_3 &= &(-3)^2 (0)^2 - 4(2)(0)^3 - 4(-3)^3 (p) - 27(2)^2 (p)^2 + 18(2)(-3)(0)(p) \\ &=& \cancel{(-3)^2 (0)^2} -\cancel{ 4(2)(0)^3} - 4(-3)^3 (p) - 27(2)^2 (p)^2 + \cancel{18(2)(-3)(0)(p) } \\ &=& -4(-3)^3 (p) - 27(2p)^2 \\ &=& 108p - 108p^2 \end{aligned}$

Since $D_3 \geq 0$ , then $108p - 108p^2 \geq 0 \Rightarrow 108p(p-1) \geq 0 \Rightarrow 0\leq p \leq 1$ . Thus the minimum value of $p$ is $\boxed0$ .

Rearranging the equation

$\Rightarrow p=3x^2-2x^3=g(x) \\ \Rightarrow g^{'}(x)=6x-6x^2 \Rightarrow g^{'}(x) \geq 0 \Rightarrow x \in \text{[0,1]}~~~~~~\text{{Function is increasing}}$

$\Rightarrow f(1)=1 , f(0)=0$

Now graph can easily be plotted

It can easily be seen from graph that,the equation has three real roots for $x \in [0,1]$ therefore $p \in [0,1]$ ,minimum value of $\Large{\boxed{\color{#3D99F6}{\boxed{\color{#D61F06}{p=0)}}}}}$

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Note : There is no restriction on equal roots.

Easy to see that if p is negative then the equation is giving always a negative value so cannot be 0 any negative x. So p is non negative .another simple observation for p=0 shows p=0 is least required