UPSC CDS 2 2021 Math Paper

Concept:Express the given number as a perfect square of the form $(\sqrt{a} - \sqrt{b})^2$.Explanation:Let $23 - 4\sqrt{15}$ be written as $(\sqrt{a} - \sqrt{b})^2 = a + b - 2\sqrt{ab}$.This gives $a + b = 23$ and $2\sqrt{ab} = 4\sqrt{15}$.Divide the second equation by 2: $\sqrt{ab} = 2\sqrt{15}$.Square both sides: $ab = 4 \times 15 = 60$.We need two numbers whose sum is 23 and product is 60: they are 20 and 3.Thus $23 - 4\sqrt{15} = 20 + 3 - 2\sqrt{20 \times 3}$.Rewrite as $(\sqrt{20})^2 + (\sqrt{3})^2 - 2\sqrt{20}\sqrt{3} = (\sqrt{20} - \sqrt{3})^2$.Simplify $\sqrt{20} = 2\sqrt{5}$, so $(\sqrt{20} - \sqrt{3})^2 = (2\sqrt{5} - \sqrt{3})^2$.Since squaring gives the same result, the square root can be taken as $\sqrt{3} - 2\sqrt{5}$ (the positive value).Therefore, the square root of $23 - 4\sqrt{15}$ is $\sqrt{3} - 2\sqrt{5}$.Answer:$\sqrt{3} - 2\sqrt{5}$ (Option C).