Surds and Indices Quiz

$\sqrt{72} = \sqrt{36 \times 2} = 6\sqrt{2}$. $\sqrt{32} = \sqrt{16 \times 2} = 4\sqrt{2}$. Total = $10\sqrt{2}$.

$3^{4 + (-2)} = 3^2 = 9$.

$(\sqrt{5})^2 + (\sqrt{3})^2 + 2(\sqrt{5})(\sqrt{3}) = 5 + 3 + 2\sqrt{15} = 8 + 2\sqrt{15}$.

Multiply by conjugate $(\sqrt{7}+3)$: $(\sqrt{7}+3) / (7-9) = -(\sqrt{7}+3)/2$.

$(64^{1/3})^2 = 4^2 = 16$.

Multiply by $(\sqrt{5} - \sqrt{3})$: $(5 + 3 - 2\sqrt{15}) / (5 - 3) = (8 - 2\sqrt{15}) / 2 = 4 - \sqrt{15}$.

$5 = 3 + 2$ and $6 = 3 \times 2$. So $\sqrt{5 + 2\sqrt{6}} = \sqrt{(\sqrt{3} + \sqrt{2})^2} = \sqrt{3} + \sqrt{2}$.

$2^{x+1} = 2^5 \Rightarrow x + 1 = 5 \Rightarrow x = 4$.

$(a^{-2})^2 \times (b^3)^2 = a^{-4} b^6 = b^6 / a^4$.

$2^{2x+1} = (2^3)^{x-1} = 2^{3x-3} \Rightarrow 2x+1 = 3x-3 \Rightarrow x = 4$.