# 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 ?