Question

simplify \\(\sqrt{x^4y^2}\\)

Explanation:

Step1: Recall square - root property

The square - root property states that \(\sqrt{ab}=\sqrt{a}\cdot\sqrt{b}\) (for \(a\geq0,b\geq0\)) and \(\sqrt{a^{n}} = a^{\frac{n}{2}}\) when \(a\geq0\) and \(n\) is an even positive integer. We can apply the property \(\sqrt{ab}=\sqrt{a}\cdot\sqrt{b}\) to \(\sqrt{x^{4}y^{2}}\) as follows:
\(\sqrt{x^{4}y^{2}}=\sqrt{x^{4}}\cdot\sqrt{y^{2}}\)

Step2: Simplify each square - root term

For \(\sqrt{x^{4}}\), using the formula \(\sqrt{a^{n}}=a^{\frac{n}{2}}\) with \(a = x\) and \(n = 4\), we have \(\sqrt{x^{4}}=x^{\frac{4}{2}}=x^{2}\).
For \(\sqrt{y^{2}}\), using the formula \(\sqrt{a^{n}}=a^{\frac{n}{2}}\) with \(a = y\) and \(n = 2\), we have \(\sqrt{y^{2}}=y^{\frac{2}{2}}=y\) (assuming \(x\geq0\) and \(y\geq0\) to get the principal square root).
Then, multiplying the two simplified terms together: \(\sqrt{x^{4}}\cdot\sqrt{y^{2}}=x^{2}\cdot y=x^{2}y\)

Answer:

\(x^{2}y\)