When does $nx^4+4x+3=0$ have real roots?

Question

Find all positive integers $n$ such that the equation $nx^4+4x+3=0$ has real roots.

I think the answer must also include the cases with $2$ real roots. But my main question is, how do I start? Thanks.

Answer 1

We have \begin{align} nx^4+4x+3 & = (n-1)x^4 + x^4+4x+3 = (n-1)x^4 + (x+1)^2((x-1)^2+2)\\ & = (n-1)x^4 + (x^2-1)^2 + 2(x+1)^2 \end{align} Hence, for $n \geq 2$, the function is a sum of squares and is always positive, since all three of them cannot be simultaneously zero and therefore has no real roots.

For $n=1$, we see that $x=-1$ is the only possible real root.

Answer 2

The expression is positive for large $x$ so the question is equivalent to asking whether the minimum is negative. Find this minimum (it is unique) by solving for zero derivative.

The condition that the minimum be negative is a polynomial inequality in $n$ which is not difficult to solve.