## Calculus: Early Transcendentals 8th Edition

Consider the function $f(x)=x^{10}-10x^2+5$. We see that $f(0)=0^{10}-10\cdot0^2+5=5$ and $f(1)=1^{10}-10\cdot 1^2+5=1-10+5=-4.$ So we have that $f$ is continuous everywhere and changes the sign when $x$ goes from $0$ to $1$. This means that there is at least one zero of $f$ i.e. a root of the given equation in the interval $(0,1)$, by the intermediate value theorem. Since $(0,1)$ is the subset of $(0,2)$ this means that there is also a root in $(0,2)$.