Question

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

Explanation:

Step1: Recall square - root property

The square - root property states that \(\sqrt{ab}=\sqrt{a}\cdot\sqrt{b}\) (where \(a\geq0\) and \(b\geq0\)) and \(\sqrt{a^{n}} = a^{\frac{n}{2}}\) for \(a\geq0\) and \(n\) is an integer. Also, we know that \(\sqrt{x^{4}y^{2}}=\sqrt{x^{4}}\cdot\sqrt{y^{2}}\) by the property \(\sqrt{ab}=\sqrt{a}\cdot\sqrt{b}\) (assuming \(x^{4}y^{2}\geq0\), which is true for real - valued \(x\) and \(y\) since \(x^{4}\geq0\) and \(y^{2}\geq0\) for all real \(x\) and \(y\)).

Step2: Simplify \(\sqrt{x^{4}}\) and \(\sqrt{y^{2}}\)

For \(\sqrt{x^{4}}\), using the formula \(\sqrt{a^{n}}=a^{\frac{n}{2}}\), when \(a = x\) and \(n = 4\), we have \(\sqrt{x^{4}}=x^{\frac{4}{2}}=x^{2}\).
For \(\sqrt{y^{2}}\), using the formula \(\sqrt{a^{n}}=a^{\frac{n}{2}}\), when \(a = y\) and \(n = 2\), we have \(\sqrt{y^{2}}=y^{\frac{2}{2}} = |y|\). But if we assume \(y\geq0\) (in the context of simplifying the radical expression, sometimes we consider the principal square root and assume the variables are non - negative unless stated otherwise), then \(\sqrt{y^{2}}=y\).

So, \(\sqrt{x^{4}y^{2}}=\sqrt{x^{4}}\cdot\sqrt{y^{2}}=x^{2}y\) (assuming \(y\geq0\)).

Answer:

\(x^{2}y\) (assuming \(y\geq0\))