Question 99

If $$\sqrt{10 - 2\sqrt{21}} + \sqrt{8 + 2\sqrt{15}} = \sqrt{a} + \sqrt{b}$$, where $$a$$ and $$b$$ are positive integers, then the value of $$\sqrt{ab}$$ is closest to:

Solution

$$\sqrt{10 - 2\sqrt{21}} + \sqrt{8 + 2\sqrt{15}} = \sqrt{a} + \sqrt{b}$$
$$\sqrt{(7 + 3 - 2\sqrt{7 \times 3})} + \sqrt{(5 + 3 + 2\sqrt{5 \times 3})} = \sqrt{a} + \sqrt{b}$$
$$\sqrt{(\sqrt{7} - \sqrt{3})^2} + \sqrt{(\sqrt{5} + \sqrt{3})^2} = \sqrt{a} + \sqrt{b}$$
$$\sqrt{7} - \sqrt{3} + \sqrt{5} + \sqrt{3} = \sqrt{a} + \sqrt{b}$$
$$\sqrt{7}  + \sqrt{5} = \sqrt{a} + \sqrt{b}$$
by comparing,
a = 7
b = 5
$$\sqrt{ab}$$ = $$\sqrt{7 \times 5}$$
= $$\sqrt{35}$$
= 5.9


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