Question

simplify \\(\sqrt{9y^2}\\)

Explanation:

Step1: Apply square - root property

We know that \(\sqrt{ab}=\sqrt{a}\cdot\sqrt{b}\) (where \(a\geq0\) and \(b\geq0\)). For \(\sqrt{9y^{2}}\), we can split it as \(\sqrt{9}\cdot\sqrt{y^{2}}\).

Step2: Simplify each square - root

We know that \(\sqrt{9} = 3\) because \(3\times3 = 9\). And for \(\sqrt{y^{2}}\), by the property of square - roots, \(\sqrt{x^{2}}=\vert x\vert\), so \(\sqrt{y^{2}}=\vert y\vert\). But if we assume \(y\) is a real number and we are simplifying in the context of non - negative values (or if \(y\) is non - negative), we can also write \(\sqrt{y^{2}} = y\) (when \(y\geq0\)). Since usually in basic simplification problems (unless specified otherwise) we assume the variable is such that the expression under the square root is non - negative and we can simplify \(\sqrt{y^{2}}\) as \(y\) (for non - negative \(y\)). So \(\sqrt{9}\cdot\sqrt{y^{2}}=3\times\vert y\vert\), and if \(y\geq0\), it is \(3y\).

Answer:

If \(y\geq0\), the simplified form is \(3y\); in general, the simplified form is \(3\vert y\vert\)