Question

assuming ( x ) and ( y ) are both positive, write the following expression in simplest radical form.( y^{2}sqrt{12x^{7}y^{4}} )

Explanation:

Step1: Factor the radicand

First, factor the number and variables inside the square root. We know that \(12 = 4\times3\), \(x^{7}=x^{6}\times x\), and \(y^{4}=y^{4}\) (which is a perfect square). So we can rewrite the radicand as:
\(12x^{7}y^{4}=4\times3\times x^{6}\times x\times y^{4}\)

Step2: Simplify the square root

Using the property of square roots \(\sqrt{ab}=\sqrt{a}\times\sqrt{b}\) (where \(a\) and \(b\) are non - negative) and \(\sqrt{a^{2}} = a\) (for \(a\geq0\)):
\(\sqrt{12x^{7}y^{4}}=\sqrt{4\times3\times x^{6}\times x\times y^{4}}=\sqrt{4}\times\sqrt{3}\times\sqrt{x^{6}}\times\sqrt{x}\times\sqrt{y^{4}}\)
Since \(\sqrt{4} = 2\), \(\sqrt{x^{6}}=x^{3}\) (because \((x^{3})^{2}=x^{6}\)) and \(\sqrt{y^{4}} = y^{2}\) (because \((y^{2})^{2}=y^{4}\)), we have:
\(\sqrt{12x^{7}y^{4}}=2\times\sqrt{3}\times x^{3}\times\sqrt{x}\times y^{2}=2x^{3}y^{2}\sqrt{3x}\)

Step3: Multiply by the outside term

Now, we multiply the outside term \(y^{2}\) with the simplified square root:
\(y^{2}\sqrt{12x^{7}y^{4}}=y^{2}\times2x^{3}y^{2}\sqrt{3x}\)
Using the rule of exponents \(a^{m}\times a^{n}=a^{m + n}\) for the \(y\) - terms (\(y^{2}\times y^{2}=y^{2 + 2}=y^{4}\)):
\(y^{2}\times2x^{3}y^{2}\sqrt{3x}=2x^{3}y^{4}\sqrt{3x}\)

Answer:

\(2x^{3}y^{4}\sqrt{3x}\)