To find the sum of \(\sqrt{4}\) and \(2\sqrt{8}\), let's compute each term separately.
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Calculate \(\sqrt{4}\): \[ \sqrt{4} = 2 \]
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Calculate \(2\sqrt{8}\): We first simplify \(\sqrt{8}\): \[ \sqrt{8} = \sqrt{4 \cdot 2} = \sqrt{4} \cdot \sqrt{2} = 2\sqrt{2} \] Now, substitute back into \(2\sqrt{8}\): \[ 2\sqrt{8} = 2 \cdot 2\sqrt{2} = 4\sqrt{2} \]
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Now, we sum \(\sqrt{4}\) and \(2\sqrt{8}\): \[ \sqrt{4} + 2\sqrt{8} = 2 + 4\sqrt{2} \]
Thus, the sum in simplified form is: \[ 2 + 4\sqrt{2} \]
Next, we determine if the result \(2 + 4\sqrt{2}\) is rational or irrational.
The number \(4\sqrt{2}\) is irrational, because \(\sqrt{2}\) is an irrational number, and multiplying an irrational number by a non-zero rational number (in this case, 4) still results in an irrational number. The sum of a rational number (2) and an irrational number (\(4\sqrt{2}\)) is always irrational.
Thus, the final answer is: \[ \text{The sum is } 2 + 4\sqrt{2}, \text{ which is irrational.} \]