Question

write each product form. see examples 33. \\(sqrt{72} \cdot sqrt{63}\\)

Explanation:

Step1: Use property of square roots

Recall that \(\sqrt{a}\cdot\sqrt{b}=\sqrt{ab}\) for \(a\geq0,b\geq0\). So, \(\sqrt{72}\cdot\sqrt{63}=\sqrt{72\times63}\).

Step2: Factor the product inside the square root

Factor 72 and 63: \(72 = 36\times2\) and \(63 = 9\times7\). Then \(72\times63=(36\times2)\times(9\times7)\).
Using associative property of multiplication, we can rewrite it as \(36\times9\times2\times7\).
Now, \(36 = 6^2\), \(9 = 3^2\), so \(36\times9 = 6^2\times3^2=(6\times3)^2 = 18^2\), and \(2\times7 = 14\).
Thus, \(72\times63 = 18^2\times14\), so \(\sqrt{72\times63}=\sqrt{18^2\times14}\).

Step3: Simplify the square root

Using the property \(\sqrt{ab}=\sqrt{a}\cdot\sqrt{b}\) again (with \(a = 18^2\) and \(b = 14\)), we get \(\sqrt{18^2\times14}=\sqrt{18^2}\cdot\sqrt{14}=18\sqrt{14}\).

Answer:

\(18\sqrt{14}\)