JAMB 1990 · UME · Q8✱✱

Simplify $\sqrt{160r^2 + \sqrt{71r^4 + \sqrt{100r^8}}}$ .

Worked solution (try it first)

Work from the innermost root: $\sqrt{100r^8} = 10r^4$ .
Next root: $\sqrt{71r^4 + 10r^4} = \sqrt{81r^4}$ , which is $9r^2$ .
Outer root: $\sqrt{160r^2 + 9r^2} = \sqrt{169r^2}$ , which is $13r$ .
So option C.

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