Coefficient of a quartic polynomial given the conditions it's a natural number and has atleast one real root

Question

Let $y=f(x)$ be a curve given by the solution of differential equation $dy/dx=4px^3-12x-8$ , $p ∈ N$, $f(0) = 24$ and $y=f(x)$ cuts x-axis at atleast one point. Find number of value(s) of p and sum of real roots of the equation $f(x) = 0$

I got that $f(x) = px^4-6x^2-8x+24$, but I don't know how to proceed. I tried assuming $α$ as the root and then finding p in terms of $α$ and then the condition $p≥1$. But couldn't proceed further. The answer given was p has one possible value and sum of real roots is $4$.

Answer 1

Often rewriting polynomials as sum of squares helps. In this case, we may write $$f(x) = (x^2-4)^2+2(x-2)^2+(p-1)x^4$$

Now if $p>1$, clearly this is positive and there is no real root.
If $p=1$, we get $x=2$ as a double root of $f(x) = (x-2)^2((x+2)^2+2)$ and sum of real roots is $4$, counting multiplicity.

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P.S. If your favourite definition of natural numbers allows $p=0$, then we get two more real roots for the resulting quadratic $f(x) = -2(3x^2+4x-12)$, whose sum is $-\frac43$ by Vieta.

Answer 2

Here's a crude solution:

• On the interval $[0, 1]$, we have that $0 \le x^4, x2, x \le 1$, so that $f(x)$ is between $0 - 6 - 8 + 24 = 10$ and $p - 0 - 0 + 24 = p+24$; because $p$ is a natural number, this latter value is at least 24. Summary: $f$ is nonzero on $[0, 1]$, so if we care about roots, we need only look at $x > 1$.
• On $[1, \infty)$, we have $x^4 \ge x^2 \ge x \ge 1$, so a similar estimate shows that $f(x) = px^4 - 6x^2 - 8x + 24 \ge px^4 - 6x^4 - 8 x^4 + 24 = (p - 14)x^4 +24$. If $p \ge 14$, this is always positive. Summary: if $f$ has a root, then $p$ must be between $0$ and $14$ (or between $1$ and $14$, depending on your definition of 'natural number').

Now what remains is 13 individual cases to be examined, which is tedious enough that I won't do it.

For the "sum of the roots" ... I don't have a good (or even bad) answer.