The sum of the first 8 terms of a GP is five times the sum of its...

Let the first term of the GP be $a$ and the common ratio be $r$.

The sum of the first $n$ terms is given by $S_n = \dfrac{a(r^n - 1)}{r - 1}$.

According to the given condition, $S_8 = 5S_4$.

$\dfrac{a(r^8 - 1)}{r - 1} = 5 \dfrac{a(r^4 - 1)}{r - 1}$

For a non-zero first term $a$ and given $r \neq 1$:

$r^8 - 1 = 5(r^4 - 1)$

$(r^4 - 1)(r^4 + 1) = 5(r^4 - 1)$

Assuming $r^4 - 1 \neq 0$ to avoid the trivial zero-sum case:

$r^4 + 1 = 5$

$r^4 = 4$

Since $r$ is a real number, $r^2$ must be positive.

$r^2 = 2$

$r = \pm\sqrt{2}$

Therefore, there are two possible real values of $r$.