## Precalculus (6th Edition) Blitzer

$2x^2\sqrt{5}$
RECALL: (i) For any non-negative real numbers a and b, $\sqrt{\dfrac{a}{b}}=\dfrac{\sqrt{a}}{\sqrt{b}}$ and $\dfrac{\sqrt{a}}{\sqrt{b}} = \sqrt{\dfrac{a}{b}}$. (ii) For any non-negative real number a, $\sqrt{a^2}=a$. (iii) $\dfrac{a^m}{a^n} = a^{m-n}, a\ne0$. (iv) $\sqrt{ab} = \sqrt{a} \cdot \sqrt{b}$. Use rule (i) above to obtain $\require{cancel} \sqrt{\dfrac{200x^3}{10x^{-1}}} \\=\sqrt{\dfrac{\cancel{200}20x^3}{\cancel{10}x^{-1}}}.$ Use rule (iii) above to obtain $\sqrt{20x^{3-(-1)}} \\=\sqrt{20x^{3+1}} \\=\sqrt{20x^4}.$ Factor the radicand so that one factor is a perfect square to obtain $\sqrt{4x^4(5)} \\=\sqrt{(2x^2)^2(5)}.$ Use rule (iv) above to obtain $\\\sqrt{(2x^2)^2} \cdot \sqrt{5}.$ Simplify to obtain $2x^2\sqrt{5}.$