Prove an equation has exactly two real roots

Question

If i want to prove and equation has exactly two real roots, how would i do so? What theorem would i use, Rolle's or Bolzano's or something else?

Answer 1

About equation $2^x+x^2=2.$

Let $f(x)=2^x+x^2$.

Hence, $f''(x)=2^x\ln^22+2>0$, which says that $f$ is a convex function.

Thus, the equation $f(x)=2$ has two roots maximum and indeed, $0.653...$ and $-1.25...$ are roots.

Answer 2

In principle, I would use Bolzano's theorem, the method consists in the fact that the function $f$ is easy to evaluate in points of my domain so that a change of sign is noticed. It is, in general, more comfortable to work with the derivative of a function, so I would also try to see if $f'$ has a root, although this is more likely to serve me only to prove existence since Rolle's theorem does not relate the roots of $f'$ with $f$. For example, for the equation $2^x = 2+x^2$ we can write it in the form $$ f(x)=0,\quad \text{con} ~ f(x)=2^x+x^2-2. $$ Then, we evaluate $f$ at $x = -2,0,1$ and we obtain that $$ f(-2)>0,\quad f(0)<0, \quad f(1)>0, $$ therefore we have that $f$ admits roots within the intervals $]-2,0[,~]0,1[$. And so, we can be even more careful in our analysis, we can also look at the graphs of the functions $g(x)=2^x$ and $h(x)=2-x^2$ (for which we can help the information given by the first and second derivatives) looking for their intersections.\ The Rolle theorem guarantees that if $f$ has $k$ roots then $f'$ has $k-1$ roots. In our example, if $f(x)=2^x+x^2-2$ had at least 3 roots then $f'$ would have at least 2 roots. Calculated $$ f'(x) = 2^x \ln(2) + 2x, $$ only has one root. This tells us that $f(x)=2^x+x^2-2$ has at most 2 roots.