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$Ahmed is solving the following equation for x. 2root(3)(x-7)+11=3 His first few steps are given below. {:[2root(3)(x-7)=-8],[root(3)(x-7)=-4],[(root(3)(x-7))^(3)=(-4)^(3)],[x-7=-64]:} Is it necessary for Ahmed to check his answers for extraneous solutions? Choose 1 answer: (A) Yes (B) No$

Ahmed is solving the following equation for $x$ . $\newline$ $2 \sqrt[3]{x-7}+11=3$ $\newline$ His first few steps are given below. $\newline$ $\begin{aligned} 2 \sqrt[3]{x-7} & =-8 \\ \sqrt[3]{x-7} & =-4 \\ (\sqrt[3]{x-7})^{3} & =(-4)^{3} \\ x-7 & =-64 \end{aligned}$ $\newline$ Is it necessary for Ahmed to check his answers for extraneous solutions? $\newline$ Choose $1$ answer: $\newline$ (A) Yes $\newline$ (B) No

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Math Problems

Algebra 1

Solve a system of equations using any method: word problems

Full solution

Q. Ahmed is solving the following equation for

x

\newline

2 \sqrt[3]{x-7}+11=3

\newline

His first few steps are given below.

\newline

\begin{aligned} 2 \sqrt[3]{x-7} & =-8 \\ \sqrt[3]{x-7} & =-4 \\ (\sqrt[3]{x-7})^{3} & =(-4)^{3} \\ x-7 & =-64 \end{aligned}

\newline

Is it necessary for Ahmed to check his answers for extraneous solutions?

\newline

Choose

1

answer:

\newline

(A) Yes

\newline

(B) No

Starting equation: Ahmed starts with the equation: $\newline$ $2\sqrt[3]{x-7} + 11 = 3$ $\newline$ He wants to isolate the term with the cube root, so he subtracts $11$ from both sides: $\newline$ $2\sqrt[3]{x-7} = 3 - 11$ $\newline$ $2\sqrt[3]{x-7} = -8$
Isolating the cube root term: Next, Ahmed divides both sides by $2$ to solve for the cube root of $(x-7)$ : $\sqrt[3]{x-7} = \frac{-8}{2}$ $\sqrt[3]{x-7} = -4$
Dividing both sides: Ahmed then cubes both sides to eliminate the cube root: $\newline$ $(\sqrt[3]{x-7})^3 = (-4)^3$ $\newline$ $x - 7 = -64$
Cubing both sides: Finally, Ahmed adds $7$ to both sides to solve for $x$ : $\newline$ $x = -64 + 7$ $\newline$ $x = -57$
Solving for $x$ : Now, regarding the question of whether it is necessary to check for extraneous solutions, the answer is yes. When dealing with equations that involve roots, especially when both sides of the equation are manipulated algebraically (like squaring or cubing), it is possible to introduce solutions that do not actually satisfy the original equation. Therefore, Ahmed should check his solution by substituting $x$ back into the original equation to ensure it is not extraneous.