Multiply Square Roots: √2 × √3 × √1 × √4 × √5 × √6 Step-by-Step

The product rule for square roots says $\sqrt{a} \cdot \sqrt{b} = \sqrt{ab}$. This works for any number of square roots, so $\sqrt{2} \cdot \sqrt{3} \cdot \sqrt{1} \cdot \sqrt{4} \cdot \sqrt{5} \cdot \sqrt{6} = \sqrt{2 \cdot 3 \cdot 1 \cdot 4 \cdot 5 \cdot 6}$!

Use prime factorization! Break 720 into prime factors: $720 = 2^4 \cdot 3^2 \cdot 5$. Then extract perfect squares: $\sqrt{2^4 \cdot 3^2 \cdot 5} = 2^2 \cdot 3 \cdot \sqrt{5} = 12\sqrt{5}$.

Let me recalculate! $720 = 2^4 \cdot 3^2 \cdot 5 = 16 \cdot 9 \cdot 5$. So $\sqrt{720} = \sqrt{16 \cdot 45} = 4\sqrt{45}$. Since $45 = 9 \cdot 5$, we get $4\sqrt{9 \cdot 5} = 4 \cdot 3\sqrt{5} = 12\sqrt{5}$. But wait - let me check the original calculation again...

Sure! Step 1: $2 \cdot 3 \cdot 1 \cdot 4 \cdot 5 \cdot 6 = 720$
Step 2: $720 = 8 \cdot 90 = 8 \cdot 9 \cdot 10 = 2^3 \cdot 3^2 \cdot 2 \cdot 5 = 2^4 \cdot 3^2 \cdot 5$
Step 3: $\sqrt{720} = \sqrt{2^4 \cdot 3^2 \cdot 5} = 2^2 \cdot 3 \cdot \sqrt{5} = 12\sqrt{5}$

Square your answer to verify! If you got $12\sqrt{5}$, then $(12\sqrt{5})^2 = 144 \cdot 5 = 720$. This matches our original product under the square root, so we're correct! ✓