Question

I need help on this question. Thank you so much. This is non-graded.

Answer 1

Answer: $Mallory\text{ made a mistake when simplifying the square root of y}^7\text{ \lparen3rd option\rparen}$

Explanation:

Given:

$\begin{gathered} Jessica:\text{ }\sqrt{4x^9y^6z} \\ Mallory:\text{ }\sqrt{9x^2y^7z^4} \end{gathered}$

To find:

which student made an error in their simplification of the radical expression and what error they made

To determine the error made, we will solve each of the radial expressions and compare our result with Jessica and Mallory

Jessica:

$\begin{gathered} \sqrt{4x^9y^6z}=\sqrt{2^2x^9(y^3)^2z}\text{ } \\ =\text{ 2y}^3(\sqrt{x^9\times z})\text{ = 2y}^3(\sqrt{x^8\times x\times z}) \\ =\text{ 2y}^3(\sqrt{(x^4)^2x\times z})\text{ = 2y}^3x^4(\sqrt{x\times z}) \\ =\text{2y}^3x^4\sqrt{xz} \\ This\text{ is the same as Jessica's result }(\text{2y})^3x^4xz \end{gathered}$

Mallory:

$\begin{gathered} \sqrt{9x^2y^7z^4}=\sqrt{3^2x^2y^7(z^2)}^2 \\ \sqrt{9x^2y^7z^4}\text{ = 3xz}^2\sqrt{y^7} \\ =3xz^2\sqrt{y^6\times y}\text{ = 3xz}^2\sqrt{(y^3)^2\times y} \\ =3xz^2y^3\sqrt{y} \\ This\text{ is different from Mallory's result 3xy}^6x^2\sqrt{y} \end{gathered}$$Mallory\text{ made a mistake when simplifying the square root of y}^7\text{ \lparen3rd option\rparen}$