algebra precalculus – $(2^{x^{2}-1} – 8)(sqrt[4]{1-5x}) = 0$ Solve the equation.

$(2^{x^{2}-1} – 8)(sqrt(4){1-5x}) = 0$ Solve the equation and find the biggest root.
I understand that either $(2^{x^{2}-1} – 8)$ or $(sqrt(4){1-5x})$ shoud be equal to $0$. Solving this further I get that $x=2$ or $x=-2$ or $x= frac{1}{5}$.However, the answer in my book says that the answer is $sqrt{3}$. Could you please, clarify ?