Let $D = \left \{ (a,b,c,d,e) : ax^4+bx^3+cx^2+dx+e=0 \text{ has a real root} \right \}$
Show that (1,2,-4,3,-2) is an interior point in $D$.
My only idea was trying to somehow use the implicit function theorem, but I can't find how.
Hints appreciated.
Here is a hint: is there any way you can take advantage of the structure of the polynomial, specifically with the coefficients you have been given to show that polynomials with close coefficients also have real roots?
ANSWER BELOW:
First we notice that $c = (1,2,-4,3,-2)$ is an element of $D$. To show that it's an interior point, we need to find an open ball around $c$ where each element of the ball is in $D$.
So let $B=B_{1/2}(c)$ be the open ball around $c$ of radius $1/2$. Then a point $d \in B$ may be written:
$$d = c+b = (1+b_1, 2+b_2, -4+b_3, 3+b_4, -2+b_5)$$
where $||b|| < 1/2$. Now notice that each component of $b$ can only be of magnitude less than $1/2$, so the fifth component of $d$ will always be negative. Moreover, the first component of $d$ will always be positive. Then any polynomial $d(x)$ with coefficient vector $d$ will have a negative value at $0$ and a positive value for large enough $x$. As polynomials are continuous, this implies that $d(x)$ has a real root.
Then every $d \in B$ is in $D$, hence we have an open ball centered at $c$ entirely within $D$, therefore $c$ is interior to $D$.