Question

simplify. remove all perfect squares from inside the square roots. assume a and b are positive. \\(sqrt{42a^{4}b^{6}} = \\)

Explanation:

Step1: Factor the radicand

We can factor \(42a^{4}b^{6}\) into the product of a perfect square and a non - perfect square factor. We know that \(a^{4}=(a^{2})^{2}\) and \(b^{6}=(b^{3})^{2}\), and \(42 = 6\times7\) (where 42 has no perfect square factors other than 1). So, \(\sqrt{42a^{4}b^{6}}=\sqrt{42\times(a^{2})^{2}\times(b^{3})^{2}}\)

Step2: Use the property of square roots \(\sqrt{xy}=\sqrt{x}\times\sqrt{y}\) (\(x\geq0,y\geq0\))

According to the property \(\sqrt{xy}=\sqrt{x}\cdot\sqrt{y}\) for \(x = 42\), \(y=(a^{2})^{2}(b^{3})^{2}\), we have \(\sqrt{42\times(a^{2})^{2}\times(b^{3})^{2}}=\sqrt{42}\times\sqrt{(a^{2})^{2}}\times\sqrt{(b^{3})^{2}}\)

Step3: Simplify the square roots of perfect squares

Since \(\sqrt{(a^{2})^{2}}=a^{2}\) (because \(a>0\)) and \(\sqrt{(b^{3})^{2}} = b^{3}\) (because \(b > 0\)), then \(\sqrt{42}\times\sqrt{(a^{2})^{2}}\times\sqrt{(b^{3})^{2}}=a^{2}b^{3}\sqrt{42}\)

Answer:

\(a^{2}b^{3}\sqrt{42}\)