Question 10
If $$\sqrt{12 + \sqrt[3]{x}} = \frac{7}{2}$$ and $$x = \frac{p}{q}$$, $$p, q$$ are natural numbers with G.C.D. $$(p, q) = 1$$, then $$p + q$$ is
Solution
Squaring both sides gives $$12 + \sqrt[3]{x} = \frac{49}{4}$$, so $$\sqrt[3]{x} = \frac{49}{4} - 12 = \frac{1}{4}$$. Cubing gives $$x = \frac{1}{64}$$, so $$p = 1$$ and $$q = 64$$, which are coprime. Hence $$p + q = 65$$.
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