If $x_i$ is the real root of the polynomial $P(x)=x^7+x-8$, then...

Question

$x_i \notin \langle 1; 2\rangle$

$x_i \in \langle \frac{8}{7}; \frac{4}{3}\rangle$

$x_i \in \langle \frac{4}{3}; \frac{3}{2}\rangle$

$x_i \in \langle \frac{3}{2}; 2\rangle$

$x_i \in \langle \frac{65}{64}; \frac{8}{7}\rangle$

Source: Lumbreras Editor

Answer 1

The polynomial $P(x)=x^7+x-8$ is clearly increasing. Consequently, since $P(1)=1+1-8=-6$, the (unique) real root is greater than $1$. We see that

$$x\gt1\implies x^7=8-x\lt8-1=7\implies x\lt7^{1/7}\approx1.320469\lt4/3$$

By the same reasoning,

$$x\lt4/3\implies x^7=8-x\gt8-4/3=20/3\implies x\gt(20/3)^{1/7}\approx1.3112975\gt8/7$$

Thus the real root is in the interval $(8/7,4/3)$.

Remark: If you don't have easy access to a calculator that evaluates seventh roots, the problem becomes much more difficult.