Question

$$5 \sqrt{ 8-4 \sqrt{ 18+3 \sqrt{ 2-2 \sqrt{ 32 } } } }$$

Answer

5*2*sqrt(2-sqrt(3*(6+sqrt(2*(1-4*sqrt(2))))))

Solution


Simplify  \(\sqrt{32}\)  to  \(4\sqrt{2}\).
\[5\sqrt{8-4\sqrt{18+3\sqrt{2-2\times 4\sqrt{2}}}}\]
Simplify  \(2\times 4\sqrt{2}\)  to  \(8\sqrt{2}\).
\[5\sqrt{8-4\sqrt{18+3\sqrt{2-8\sqrt{2}}}}\]
Factor out the common term \(2\).
\[5\sqrt{8-4\sqrt{18+3\sqrt{2(1-4\sqrt{2})}}}\]
Factor out the common term \(3\).
\[5\sqrt{8-4\sqrt{3(6+\sqrt{2(1-4\sqrt{2})})}}\]
Factor out the common term \(4\).
\[5\sqrt{4(2-\sqrt{3(6+\sqrt{2(1-4\sqrt{2})})})}\]
Use this rule: \(\sqrt{ab}=\sqrt{a}\sqrt{b}\).
\[5\sqrt{4}\sqrt{2-\sqrt{3(6+\sqrt{2(1-4\sqrt{2})})}}\]
Since \(2\times 2=4\), the square root of \(4\) is \(2\).
\[5\times 2\sqrt{2-\sqrt{3(6+\sqrt{2(1-4\sqrt{2})})}}\]