The square root of $$33-4\sqrt{35}$$ is :
To find : $$\sqrt{33-4\sqrt{35}}$$
We can write it as :
= $$\sqrt{33 - 2 * 2 * \sqrt{7} * \sqrt{5}}$$
Since, $$(a^2 + b^2 - 2ab) = (a-b)^2$$
= $$\sqrt{(2\sqrt{7})^2 + (\sqrt{5})^2 - 2*2\sqrt{7}*\sqrt{5}}$$
= $$\sqrt{(2\sqrt{7} - \sqrt{5})^2}$$
= $$\pm(2\sqrt{7}-\sqrt{5})$$
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