Proving the equation $4x^3+6x^2+5x=-7$ has has only one solution using Rolle's or Lagrange's theorem

Question

Prove the equation $4x^3+6x^2+5x=-7$ has only one solution.

Hello all. It is easy to prove this by defining the function $f(x)=4x^3+6x^2+5x+7$ and claiming it is a monotonic increasing function at $\mathbb{R}$, and since it is continuous at $\mathbb{R}$, and $f(-2)\le 0, f(-1)\ge 0$ then based on the MVT there exists a point $c\in[-2,-1]$ such that $f(c)=0$.

How do I prove there's only one real root using Rolle's theorem or Lagrange? Thanks in advance :)

Answer 1

For example, let's assume that $p(x)=4x^3+6x^2+5x+7=0$ has $2$ solutions $x_1<x_2$ such that $p(x_1)=p(x_2)=0$. Then, by Rolle's theorem, there must be $c \in (x_1,x_2)$ such that $p'(c)=0$. But $p'(x)=12x^2+12x+5=0$ has no zeros, since $\Delta=12^2-4\cdot 5\cdot 12=-96<0$.

Answer 2

If a polynomial has two distinct roots (say $a<b$) then it's derivative has a root at some $x\in (a,b)$ (why? This is where you apply the theorems you mentioned). But it easy to see that the derivative has no root, hence $f$ at most one. That such a root exists follows from its behavior at $\pm \infty$

Answer 3

Assume it has two roots.

$$f (a)=f (b)=0$$

then by Rolle's Theorem there will exist $c\in (a,b) $ such that

$$f'(c)=0=12c^2+12c+5$$

but the reduced discriminant is $$\delta'=36-60 <0$$

Done.