Question

# Solve  $$\dfrac { 1 }{ \sqrt { 19-\sqrt { 360 } } } -\dfrac { 1 }{ \sqrt { 21-\sqrt { 440 } } } +\dfrac { 2 }{ \sqrt { 20+\sqrt { 396 } } } =?$$

Solution

## Consider the first term$$\Rightarrow \dfrac{1}{\sqrt {19-\sqrt {360}}}$$On rationalizing, we get$$\Rightarrow \dfrac{\sqrt {19+\sqrt {360}}}{\sqrt {19^2-360}}=\sqrt {19+\sqrt {360}}$$$$\Rightarrow \sqrt{19+\sqrt{360}}=\sqrt{19+2\sqrt{90}}=\sqrt{9+10+2\sqrt{9}\sqrt {10}}=\sqrt{(\sqrt 9+\sqrt 10)^2}=\sqrt 9+\sqrt {10}$$Now consider the second term$$\Rightarrow \dfrac{1}{\sqrt {21-\sqrt {440}}}$$On rationalizing, we get$$\Rightarrow \dfrac{\sqrt {21+\sqrt {440}}}{\sqrt {21^2-440}}=\sqrt {21+\sqrt {440}}$$$$\Rightarrow \sqrt{21+\sqrt{440}}=\sqrt{21+2\sqrt{110}}=\sqrt{10+11+2\sqrt{10}\sqrt {11}}=\sqrt{(\sqrt {10}+\sqrt {11})^2}=\sqrt {10}+\sqrt {11}$$Now consider the third term$$\Rightarrow \dfrac{2}{\sqrt {20-\sqrt {396}}}$$On rationalizing, we get$$\Rightarrow \dfrac{2\sqrt {20-\sqrt {396}}}{\sqrt {20^2-396}}=\sqrt {20-\sqrt {396}}$$$$\Rightarrow \sqrt{20-\sqrt{396}}=\sqrt{20-2\sqrt{99}}=\sqrt{9+11-2\sqrt{9}\sqrt {11}}=\sqrt{(\sqrt {11}-\sqrt {9})^2}=\sqrt {11}-\sqrt {9}$$Substituting these values in the given expression we get,$$\Rightarrow (\sqrt 9+\sqrt {10})-(\sqrt {10}+\sqrt {11})+(\sqrt {11}-\sqrt {9})$$$$\Rightarrow 0$$Mathematics

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