Given that \(\sqrt{128}+\sqrt{18}-\sqrt{K} = 7\sqrt{2}\), find K,
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Correct Answer: Option C
Explanation:
\(\sqrt{128} +\sqrt{18}-\sqrt{k}=7\sqrt{2}\\
\sqrt{2\times 64}+\sqrt{9\times 2} - \sqrt{k} = 7\sqrt{2}\\
8\sqrt{2} + 3\sqrt{2} - \sqrt{k} = 7\sqrt{2}; 11\sqrt{2} - \sqrt{k} = 7\sqrt{2}\\
-\sqrt{k}=7\sqrt{2}-11\sqrt{2}; -\sqrt{k} = -4\sqrt{2}; \sqrt{k}=4\sqrt{2}\\
=\sqrt{4^2\times 2} = \sqrt{16\times 2}; \sqrt{k}=\sqrt{32}; k = 32\)
\(\sqrt{128} +\sqrt{18}-\sqrt{k}=7\sqrt{2}\\
\sqrt{2\times 64}+\sqrt{9\times 2} - \sqrt{k} = 7\sqrt{2}\\
8\sqrt{2} + 3\sqrt{2} - \sqrt{k} = 7\sqrt{2}; 11\sqrt{2} - \sqrt{k} = 7\sqrt{2}\\
-\sqrt{k}=7\sqrt{2}-11\sqrt{2}; -\sqrt{k} = -4\sqrt{2}; \sqrt{k}=4\sqrt{2}\\
=\sqrt{4^2\times 2} = \sqrt{16\times 2}; \sqrt{k}=\sqrt{32}; k = 32\)