simplify using prime factorization trees and the index.
if $sqrtn{x^n}$ comes out with odd exponent (where n is even), you must use $|x|$
11) $sqrt{18x^2y^2}$
12) $(64x^3y^5)^{\frac{1}{4}}$

Question

Accepted Answer

### Problem 11: $\boldsymbol{\sqrt{18x^2y^2}}$
# Explanation:
## Step 1: Prime Factorize 18
Prime factorize 18: $18 = 2	imes9 = 2	imes3^2$. So, $\sqrt{18x^2y^2}=\sqrt{2	imes3^2	imes x^2	imes y^2}$.

## Step 2: Apply Square Root Property
For a square root (index 2, even), we use the property $\sqrt{a^2}=|a|$ if the exponent of $a$ in the radicand is odd, and $\sqrt{a^2}=a$ if even. Here, exponents of 3, $x$, $y$ are 2 (even), exponent of 2 is 1 (odd).
So, $\sqrt{3^2}=3$, $\sqrt{x^2}=|x|$, $\sqrt{y^2}=|y|$, and $\sqrt{2}$ remains as is.
Combining these, we get $3|x||y|\sqrt{2}$. Since multiplication of absolute values is commutative, we can write it as $3\sqrt{2}|x||y|$ or $3|x||y|\sqrt{2}$.

# Answer:
$3\sqrt{2}|x||y|$

### Problem 12: $\boldsymbol{(64x^3y^5)^{\frac{1}{4}}}$
# Explanation:
## Step 1: Prime Factorize 64
Prime factorize 64: $64 = 2^6$. So, $(64x^3y^5)^{\frac{1}{4}}=(2^6x^3y^5)^{\frac{1}{4}}$.

## Step 2: Apply Exponent Rule $(a^m b^n)^p = a^{mp}b^{np}$
Apply the exponent rule:
- For 2: $2^{6	imes\frac{1}{4}} = 2^{\frac{6}{4}} = 2^{\frac{3}{2}}=\sqrt{2^3}=2\sqrt{2}$ (or keep as $2^{\frac{3}{2}}$).
- For $x$: $x^{3	imes\frac{1}{4}} = x^{\frac{3}{4}}=\sqrt[4]{x^3}$.
- For $y$: $y^{5	imes\frac{1}{4}} = y^{\frac{5}{4}} = y^{\frac{4 + 1}{4}} = y^{1+\frac{1}{4}} = y\sqrt[4]{y}$.

## Step 3: Combine Terms
Combining the results for 2, $x$, and $y$:
$2^{\frac{3}{2}}x^{\frac{3}{4}}y^{\frac{5}{4}} = 2\sqrt{2}\cdot\sqrt[4]{x^3}\cdot y\sqrt[4]{y}=2y\sqrt{2}\cdot\sqrt[4]{x^3y}$.
Or, in radical form without simplifying the fractional exponents further:
$(2^6x^3y^5)^{\frac{1}{4}} = 2^{\frac{6}{4}}x^{\frac{3}{4}}y^{\frac{5}{4}} = 2^{\frac{3}{2}}x^{\frac{3}{4}}y^{\frac{5}{4}} = \sqrt{2^3} \cdot \sqrt[4]{x^3} \cdot \sqrt[4]{y^5} = 2\sqrt{2} \cdot \sqrt[4]{x^3} \cdot y\sqrt[4]{y} = 2y\sqrt{2}\sqrt[4]{x^3y}$.

Alternatively, using absolute value for even index (4 is even) when exponents are odd in the radicand (but here we have fractional exponents, so we can express in radical form with absolute value if needed, but since the original exponent of $x$ is 3 (odd) and $y$ is 5 (odd) in the base, when raising to $\frac{1}{4}$ (even index equivalent in radical), we need to check:

Wait, actually, the rule is for $\sqrt[n]{x^m}$ where $n$ is even. Here, $(a)^{\frac{1}{4}}=\sqrt[4]{a}$, so $a = x^3y^5$. The exponents of $x$ (3) and $y$ (5) in $a$: when taking 4th root (even index), for $x^3$, since 3 is odd, $\sqrt[4]{x^3}=|x|^{\frac{3}{4}}$? Wait, no, the correct property is $\sqrt[n]{x^m}=|x|^{\frac{m}{n}}$ if $n$ is even and $m$ is odd (because $x^m$ under even root: to make it defined for all real $x$, we use absolute value when the exponent of $x$ in the radicand is odd, and the index is even). Wait, actually, for real numbers, if we consider $x$ and $y$ as real numbers, then:

- $\sqrt[4]{x^3}=|x|^{\frac{3}{4}}$ (since 4 is even, 3 is odd: $\sqrt[4]{x^3}=|x|^{\frac{3}{4}}$ because $x^3$ can be negative if $x$ is negative, and 4th root of negative is not real, so we need $|x|^{\frac{3}{4}}$ to ensure it's real for all real $x$? Wait, no, actually, if $x$ is negative, $x^3$ is negative, and 4th root of negative is not real. So, to have a real result, we need $x^3\geq0$ (i.e., $x\geq0$) or we use absolute value? Wait, maybe the problem assumes $x$ and $y$ are non-negative? The original problem statement says "using Prime Factorization Trees and the index" and the note about odd exponent with even $n$: maybe in the context of the problem, we can proceed with the exponent rules without absolute value if we assume $x,y\geq0$, or include absolute value for real numbers.

Assuming we need to consider real numbers (so include absolute value when necessary):

- For $x^3$: exponent 3 (odd), index 4 (even). So $\sqrt[4]{x^3}=|x|^{\frac{3}{4}}$.
- For $y^5$: exponent 5 (odd), index 4 (even). So $\sqrt[4]{y^5}=|y|^{\frac{5}{4}}=|y|^{1+\frac{1}{4}}=|y|\cdot|y|^{\frac{1}{4}}=|y|\sqrt[4]{|y|}$ (but $|y|^{\frac{5}{4}}=|y|\cdot|y|^{\frac{1}{4}}$ since $\frac{5}{4}=1+\frac{1}{4}$). Wait, maybe better to keep as exponents:

From step 2, we had:

$2^{\frac{3}{2}}x^{\frac{3}{4}}y^{\frac{5}{4}}$.

For real numbers, since 4 is even, and exponents of $x$ (3) and $y$ (5) in the base are odd, we need to write:

$2^{\frac{3}{2}}|x|^{\frac{3}{4}}|y|^{\frac{5}{4}}$.

Simplify $|y|^{\frac{5}{4}} = |y|^{1+\frac{1}{4}} = |y|\cdot|y|^{\frac{1}{4}} = |y|\sqrt[4]{|y|}$, and $2^{\frac{3}{2}} = 2\sqrt{2}$, $|x|^{\frac{3}{4}}=\sqrt[4]{|x|^3}$.

So combining:

$2\sqrt{2}\cdot\sqrt[4]{|x|^3}\cdot|y|\cdot\sqrt[4]{|y|} = 2\sqrt{2}|y|\sqrt[4]{|x|^3|y|}$.

But maybe the problem expects us to proceed with the exponent rules without absolute value (assuming $x,y\geq0$). Let's check both ways.

If we assume $x,y\geq0$ (so $|x|=x$, $|y|=y$):

From step 2:

$2^{\frac{3}{2}}x^{\frac{3}{4}}y^{\frac{5}{4}} = 2\sqrt{2}x^{\frac{3}{4}}y^{\frac{5}{4}} = 2\sqrt{2}x^{\frac{3}{4}}y y^{\frac{1}{4}} = 2\sqrt{2}y x^{\frac{3}{4}}y^{\frac{1}{4}} = 2\sqrt{2}y\sqrt[4]{x^3y}$.

So, depending on the context (with or without absolute value), but since the problem's note was about $\sqrt[n]{x^n}$ with odd exponent and even $n$, here we have a 4th root (even index) of $x^3$ (exponent 3, odd) and $y^5$ (exponent 5, odd). So, to make the expression real for all real $x,y$, we need to include absolute values:

$2^{\frac{3}{2}}|x|^{\frac{3}{4}}|y|^{\frac{5}{4}} = 2\sqrt{2}|x|^{\frac{3}{4}}|y|^{\frac{5}{4}} = 2\sqrt{2}|y|\sqrt[4]{|x|^3|y|}$.

But if we assume $x,y\geq0$, then it's $2\sqrt{2}y\sqrt[4]{x^3y}$.

Let's proceed with the exponent rules as per the problem's instruction (maybe assuming non-negative variables for simplicity, as prime factorization trees are usually for positive integers, and variables might be non-negative in this context):

So, $(64x^3y^5)^{\frac{1}{4}} = (2^6x^3y^5)^{\frac{1}{4}} = 2^{\frac{6}{4}}x^{\frac{3}{4}}y^{\frac{5}{4}} = 2^{\frac{3}{2}}x^{\frac{3}{4}}y^{\frac{5}{4}} = 2\sqrt{2}x^{\frac{3}{4}}y^{\frac{5}{4}} = 2\sqrt{2}y\sqrt[4]{x^3y}$ (since $y^{\frac{5}{4}} = y\cdot y^{\frac{1}{4}} = y\sqrt[4]{y}$ and $x^{\frac{3}{4}}=\sqrt[4]{x^3}$).

# Answer:
$2\sqrt{2}y\sqrt[4]{x^3y}$ (or $2^{\frac{3}{2}}x^{\frac{3}{4}}y^{\frac{5}{4}}$ or with absolute values: $2\sqrt{2}|y|\sqrt[4]{|x|^3|y|}$)

Sovi.AI - AI Math Tutor

QUESTION IMAGE

Question

Explanation:

Problem 11: $\boldsymbol{\sqrt{18x^2y^2}}$

Step 1: Prime Factorize 18

Step 2: Apply Square Root Property

Step 1: Prime Factorize 64

Step 2: Apply Exponent Rule $(a^m b^n)^p = a^{mp}b^{np}$

Step 3: Combine Terms

Answer:

Problem 12: $\boldsymbol{(64x^3y^5)^{\frac{1}{4}}}$