Question

The product of 5sqrt(27) 27and sqrt(15) would yield a rational number

Answer 1

To determine if the product \( 5\sqrt{27} \), \( 27 \), and \( \sqrt{15} \) yields a rational number, we first simplify each term and then multiply them together.

1. Simplify \( \sqrt{27} \):
\[
\sqrt{27} = \sqrt{9 \times 3} = \sqrt{9} \times \sqrt{3} = 3\sqrt{3}
\]

2. Now substitute back into the product:
\[
5\sqrt{27} = 5 \times 3\sqrt{3} = 15\sqrt{3}
\]

3. Next, we multiply \( 15\sqrt{3} \) and \( 27 \):
\[
15\sqrt{3} \times 27 = 15 \times 27 \times \sqrt{3} = 405\sqrt{3}
\]

4. Now multiply \( 405\sqrt{3} \) by \( \sqrt{15} \):
\[
405\sqrt{3} \times \sqrt{15} = 405 \times \sqrt{3 \times 15} = 405 \sqrt{45}
\]

5. Simplify \( \sqrt{45} \):
\[
\sqrt{45} = \sqrt{9 \times 5} = \sqrt{9} \times \sqrt{5} = 3\sqrt{5}
\]

6. So now we have:
\[
405\sqrt{45} = 405 \times 3\sqrt{5} = 1215\sqrt{5}
\]

The final result \( 1215\sqrt{5} \) is not a rational number since it contains the irrational component \( \sqrt{5} \).

Thus, the product \( 5\sqrt{27} \times 27 \times \sqrt{15} \) does not yield a rational number.