\(-\sqrt{8}+\sqrt{128}-\sqrt{32}\) is equal to
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → \(2\sqrt{2}\)
Surds can be added or subtracted only when they are like surds. Each root is written as a multiple of \(\sqrt{2}\) by taking out the largest perfect square.
\(\sqrt{8}=\sqrt{4\times2}=2\sqrt{2}\)
\(\sqrt{128}=\sqrt{64\times2}=8\sqrt{2}\)
\(\sqrt{32}=\sqrt{16\times2}=4\sqrt{2}\)
\(-2\sqrt{2}+8\sqrt{2}-4\sqrt{2}=2\sqrt{2}\)