Given sqrt{2} = 1.414. The value of sqrt{8} + | vedclass

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(B) First,simplify each radical term:$\sqrt{8} = \sqrt{4 	imes 2} = 2\sqrt{2}$$2\sqrt{32} = 2\sqrt{16 	imes 2} = 2 	imes 4\sqrt{2} = 8\sqrt{2}$$3\s

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$8.484$

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(B) First,simplify each radical term:$\sqrt{8} = \sqrt{4 	imes 2} = 2\sqrt{2}$$2\sqrt{32} = 2\sqrt{16 	imes 2} = 2 	imes 4\sqrt{2} = 8\sqrt{2}$$3\sqrt{128} = 3\sqrt{64 	imes 2} = 3 	imes 8\sqrt{2} = 24\sqrt{2}$$4\sqrt{50} = 4\sqrt{25 	imes 2} = 4 	imes 5\sqrt{2} = 20\sqrt{2}$Now,substitute these into the expression:$2\sqrt{2} + 8\sqrt{2} - 24\sqrt{2} + 20\sqrt{2}$$= (2 + 8 - 24 + 20)\sqrt{2}$$= 6\sqrt{2}$Given $\sqrt{2} = 1.414$,we have:$6 	imes 1.414 = 8.484$

Given $\sqrt{2} = 1.414$. The value of $\sqrt{8} + 2\sqrt{32} - 3\sqrt{128} + 4\sqrt{50}$ is

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