Question

6th period alg 2 5-1 radicals/roots simplify the radical. assume that all variables represent positive real numbers.
( sqrt5{-32x^{10}y^{5}} )
select the correct choice below and, if necessary, fill in the answer box to complete your choice.
a. ( sqrt5{-32x^{10}y^{5}} = square )
b. the radical does not represent a real number.

Explanation:

Step1: Analyze the radicand for the 5th root

We know that for a 5th root (odd index), the root of a negative number is defined. Let's break down the radicand \(-32x^{10}y^{5}\). We can express each part as a perfect 5th power:

• \(-32 = (-2)^5\) (since \((-2)\times(-2)\times(-2)\times(-2)\times(-2)= -32\))
• \(x^{10}=(x^{2})^5\) (using the power - of - a - power rule \((a^{m})^{n}=a^{mn}\), here \(m = 2\), \(n = 5\), so \((x^{2})^{5}=x^{10}\))
• \(y^{5}=y^{5}\) (since \(y^{5}=(y)^{5}\))

Step2: Apply the 5th root property

The property of the \(n\)th root is \(\sqrt[n]{ab}=\sqrt[n]{a}\cdot\sqrt[n]{b}\) (for \(n\) odd, \(a\) and \(b\) real numbers) and \(\sqrt[n]{a^{n}}=a\) (for \(n\) odd).
So, \(\sqrt[5]{-32x^{10}y^{5}}=\sqrt[5]{(-2)^{5}\cdot(x^{2})^{5}\cdot y^{5}}\)
Using the property \(\sqrt[n]{ab}=\sqrt[n]{a}\cdot\sqrt[n]{b}\), we can write this as \(\sqrt[5]{(-2)^{5}}\cdot\sqrt[5]{(x^{2})^{5}}\cdot\sqrt[5]{y^{5}}\)
Then, using the property \(\sqrt[n]{a^{n}}=a\) (for \(n = 5\) odd), we get \((-2)\cdot x^{2}\cdot y\)

Step3: Simplify the expression

Multiplying the terms together, we have \(-2x^{2}y\)

Answer:

\(-2x^{2}y\) (and the correct choice is A, since the radical represents a real number and simplifies to \(-2x^{2}y\))