The product if 5sqrt27 and ___ would yield a rational number.

A. sqrt12
B. sqrt9
C. 4sqrt2
D. sqrt15

1 answer

To find which option makes the product of \( 5\sqrt{27} \) with it a rational number, we start by simplifying \( 5\sqrt{27} \).

We can rewrite \( \sqrt{27} \) as \( \sqrt{9 \cdot 3} = \sqrt{9} \cdot \sqrt{3} = 3\sqrt{3} \). Thus:

\[ 5\sqrt{27} = 5 \cdot 3\sqrt{3} = 15\sqrt{3}. \]

Now we need to multiply \( 15\sqrt{3} \) by each option and check if the product is a rational number (a number that can be expressed as a fraction of integers).

  1. Option A: \( \sqrt{12} \)

\[ 15\sqrt{3} \cdot \sqrt{12} = 15\sqrt{3} \cdot \sqrt{4 \cdot 3} = 15\sqrt{3} \cdot 2\sqrt{3} = 30 \cdot 3 = 90 \] (Rational number)

  1. Option B: \( \sqrt{9} \)

\[ 15\sqrt{3} \cdot \sqrt{9} = 15\sqrt{3} \cdot 3 = 45\sqrt{3} \] (Not a rational number)

  1. Option C: \( 4\sqrt{2} \)

\[ 15\sqrt{3} \cdot 4\sqrt{2} = 60\sqrt{6} \] (Not a rational number)

  1. Option D: \( \sqrt{15} \)

\[ 15\sqrt{3} \cdot \sqrt{15} = 15\sqrt{45} = 15\sqrt{9 \cdot 5} = 15 \cdot 3\sqrt{5} = 45\sqrt{5} \] (Not a rational number)

The only option that results in a rational product is:

A. \( \sqrt{12} \).