To solve the equation \( x^2 = 27 \), we take the square root of both sides:
\[ x = \sqrt{27} \]
We can simplify \( \sqrt{27} \) as follows:
\[ \sqrt{27} = \sqrt{9 \times 3} = \sqrt{9} \cdot \sqrt{3} = 3\sqrt{3} \]
Now, we need to determine the nature of the number \( 3\sqrt{3} \).
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Irrational Number: The number \( \sqrt{3} \) is an irrational number. Therefore, any multiple of it (such as \( 3\sqrt{3} \)) is also irrational.
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Repeating Decimal: Since \( 3\sqrt{3} \) is irrational, it cannot be expressed as a repeating decimal.
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Rational Number: \( 3\sqrt{3} \) is not a rational number since it cannot be written as a fraction of two integers.
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Greater than Zero but Less Than One: Clearly, \( 3\sqrt{3} \) is greater than zero because both 3 and \( \sqrt{3} \) are positive. However, \( \sqrt{3} \) is approximately \( 1.732 \), so \( 3\sqrt{3} \) is approximately \( 3 \times 1.732 = 5.196 \), which is greater than 1, not less than 1.
Based on the explanations, the best description of the positive solution \( x = 3\sqrt{3} \) is:
A. It is an irrational number.