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Question
√(xyz²)
plify
Step1: Recall square - root property
We know that for any non - negative real number \(a\) and \(b\), \(\sqrt{ab}=\sqrt{a}\cdot\sqrt{b}\), and \(\sqrt{a^{2}}=\vert a\vert\). In the expression \(\sqrt{xyz^{2}}\), we can use the property \(\sqrt{ab}=\sqrt{a}\cdot\sqrt{b}\) to split the square root. So, \(\sqrt{xyz^{2}}=\sqrt{x}\cdot\sqrt{y}\cdot\sqrt{z^{2}}\).
Step2: Simplify \(\sqrt{z^{2}}\)
Using the property \(\sqrt{a^{2}}=\vert a\vert\), when \(a = z\), we have \(\sqrt{z^{2}}=\vert z\vert\). But if we assume that \(z\) is a non - negative real number (since in many basic algebraic simplifications we consider the principal square root and assume variables are non - negative in the domain of the square root for simplicity), then \(\vert z\vert=z\). So, \(\sqrt{xyz^{2}}=\vert z\vert\sqrt{xy}\) (or \(z\sqrt{xy}\) if \(z\geq0\)).
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If \(z\geq0\), the simplified form is \(z\sqrt{xy}\); in general, it is \(\vert z\vert\sqrt{xy}\)