Question

simplify the following expression.
\\(\sqrt{x^{5}y^{5}} \cdot 4\sqrt{2x^{13}y^{4}}\\)

Explanation:

Step1: Simplify each square root separately

First, simplify \(\sqrt{x^{5}y^{5}}\). We can rewrite the exponents as sums of even and odd parts. For \(x^{5}\), we have \(x^{4 + 1}=x^{4}\cdot x\), and for \(y^{5}\), we have \(y^{4+1}=y^{4}\cdot y\). So, \(\sqrt{x^{5}y^{5}}=\sqrt{x^{4}\cdot x\cdot y^{4}\cdot y}\). Using the property \(\sqrt{ab}=\sqrt{a}\cdot\sqrt{b}\) (where \(a = x^{4}\cdot y^{4}\) and \(b=xy\)), we get \(\sqrt{x^{4}\cdot y^{4}}\cdot\sqrt{xy}=x^{2}y^{2}\sqrt{xy}\) (since \(\sqrt{x^{4}} = x^{2}\) and \(\sqrt{y^{4}}=y^{2}\) for non - negative \(x,y\)).

Next, simplify \(4\sqrt{2x^{13}y^{4}}\). For \(x^{13}\), we can write \(x^{12 + 1}=x^{12}\cdot x\), and \(\sqrt{y^{4}}=y^{2}\). So, \(\sqrt{2x^{13}y^{4}}=\sqrt{2\cdot x^{12}\cdot x\cdot y^{4}}=\sqrt{x^{12}\cdot y^{4}}\cdot\sqrt{2x}=x^{6}y^{2}\sqrt{2x}\). Then \(4\sqrt{2x^{13}y^{4}} = 4x^{6}y^{2}\sqrt{2x}\).

Step2: Multiply the two simplified expressions

Now we multiply \(x^{2}y^{2}\sqrt{xy}\) and \(4x^{6}y^{2}\sqrt{2x}\). First, multiply the coefficients and the variables with the same base separately, and then multiply the square roots.

• Coefficients: \(1\times4 = 4\)
• For \(x\): \(x^{2}\cdot x^{6}=x^{2 + 6}=x^{8}\) (using the rule \(a^{m}\cdot a^{n}=a^{m + n}\))
• For \(y\): \(y^{2}\cdot y^{2}=y^{2+2}=y^{4}\)
• For the square roots: \(\sqrt{xy}\cdot\sqrt{2x}=\sqrt{xy\cdot2x}=\sqrt{2x^{2}y}\) (using the rule \(\sqrt{a}\cdot\sqrt{b}=\sqrt{ab}\))

So now we have \(4x^{8}y^{4}\sqrt{2x^{2}y}\).

Step3: Simplify the remaining square root

We know that \(\sqrt{x^{2}}=x\) (for non - negative \(x\)), so \(\sqrt{2x^{2}y}=x\sqrt{2y}\). Substitute this back into the expression:

\(4x^{8}y^{4}\cdot x\sqrt{2y}=4x^{8 + 1}y^{4}\sqrt{2y}=4x^{9}y^{4}\sqrt{2y}\)

Answer:

\(4x^{9}y^{4}\sqrt{2y}\)