High School

Select the correct answer from each drop-down menu.

Review Seth's steps for rewriting and simplifying an expression.

Given:
[tex]8 x^6 \sqrt{200 x^{13}} \div 2 x^5 \sqrt{32 x^7}[/tex]

Step 1:
[tex]8 x^6 \sqrt{4 \cdot 25 \cdot 2 \cdot (x^6)^2 \cdot x} \div 2 x^5 \sqrt{16 \cdot 2 \cdot (x^3)^2 \cdot x}[/tex]

Step 2:
[tex]8 \cdot 2 \cdot 5 \cdot x^6 \cdot x^6 \sqrt{2 x} \div 2 \cdot 16 \cdot x^5 \cdot x^3 \sqrt{2 x}[/tex]

Step 3:
[tex]80 x^{12} \sqrt{2 x} \div 32 x^8 \sqrt{2 x}[/tex]

Step 4:
[tex]\frac{80 x^{11} \sqrt{2 \pi}}{32 x^3 \sqrt{2 x}}[/tex]

Step 5:
[tex]\frac{5}{2} x^4[/tex]

Seth's first mistake was made in [tex]\square[/tex], where he [tex]\square[/tex].

Answer :

To identify Seth's first mistake in simplifying the expression, let's carefully go through each of his steps.

The given expression was:

[tex]\[ 8 x^6 \sqrt{200 x^{13}} \div 2 x^5 \sqrt{32 x^7} \][/tex]

Let's examine Seth's steps:

Step 1: Rewriting the square roots as:

[tex]\[ 8 x^6 \sqrt{4 \cdot 25 \cdot 2 \cdot (x^6)^2 \cdot x} \div 2 x^5 \sqrt{16 \cdot 2 \cdot (x^3)^2 \cdot x} \][/tex]

This step correctly breaks down the square roots into prime factors, though it's somewhat unnecessarily complex.

Step 2: Simplifies to:

[tex]\[ 8 \cdot 2 \cdot 5 \cdot x^6 \cdot x^6 \sqrt{2 x} \div 2 \cdot 16 \cdot x^5 \cdot x^3 \sqrt{2 x} \][/tex]

This step simplifies the square roots inside, but it looks like Seth made a mistake in the simplification here.

Step 3: Simplifies to:

[tex]\[ 80 x^{12} \sqrt{2 x} \div 32 x^8 \sqrt{2 x} \][/tex]

Here, Seth cancels out common terms, but observe that there is an inconsistency because later steps show cancellation and final values that don't result in what's expected from here.

Step 4: Further simplifies incorrectly:

[tex]\[ \frac{80 x^{11} \sqrt{2 \pi}}{32 x^3 \sqrt{2 x}} \][/tex]

Pi ([tex]\(\pi\)[/tex]) has unexpectedly appeared in the square root, indicating an error in copying or arithmetic before this.

Step 5: Final result:

[tex]\[ \frac{5}{2} x^4 \][/tex]

The first mistake happened in Step 4. The mistake was the introduction of [tex]\(\pi\)[/tex] in [tex]\(\sqrt{2 \pi}\)[/tex], which does not appear in any previous steps and is completely incorrect.

In summary, Seth's first error occurred in Step 4, where he incorrectly introduced [tex]\(\pi\)[/tex] in the square root.