Question 35

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waecmaths question: 

Simplify $\sqrt{\frac{{{8}^{2}}\times {{4}^{n+1}}}{{{2}^{2n}}\times 16}}$

Option A: 

16

Option B: 

8

Option C: 

4

Option D: 

1

waecmaths solution: 

$\begin{align}  & \sqrt{\frac{{{8}^{2}}\times {{4}^{n+1}}}{{{2}^{2n}}\times 16}}=\sqrt{\frac{{{({{2}^{3}})}^{2}}\times {{2}^{2(n+1)}}}{{{2}^{2n}}\times {{2}^{4}}}}=\sqrt{\frac{{{2}^{6}}\times {{2}^{2n+2}}}{{{2}^{2n}}\times {{2}^{4}}}} \\ & \sqrt{\frac{{{8}^{2}}\times {{4}^{n+1}}}{{{2}^{2n}}\times 16}}=\sqrt{\frac{{{2}^{8+2n}}}{{{2}^{2n+4}}}}=\sqrt{{{2}^{8+2n-2n-4}}} \\ & \sqrt{\frac{{{8}^{2}}\times {{4}^{n+1}}}{{{2}^{2n}}\times 16}}=\sqrt{{{2}^{4}}}={{({{2}^{4}})}^{\tfrac{1}{2}}}={{2}^{2}}=4 \\\end{align}$

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