Question

a student concluded that the solution to the equation \\(\sqrt{2x+1}+3=0\\) is \\(x=4\\).

do you agree? explain why or why not.

Explanation:

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<reasoning>

Substitute the proposed solution

Using the Solving Radical Equations knowledge point
\[

$$\begin{aligned} &\text{Substitute } x = 4 \text{ into the left side of the equation:}\\ &\sqrt{2(4)+1}+3 = \sqrt{8+1}+3 = \sqrt{9}+3 \end{aligned}$$

\]

Evaluate the principal square root

Using the Principal Square Root Definition knowledge point
\[

$$\begin{aligned} &\sqrt{9} = 3\\ &3 + 3 = 6 e 0 \end{aligned}$$

\]

Identify the extraneous solution

Using the Extraneous Solutions knowledge point
\[

$$\begin{aligned} &\sqrt{2x+1} = -3\\ &\text{Since } \sqrt{2x+1} \ge 0 \text{ for all real } x, \text{ there are no real solutions.}\\ &x = 4 \text{ is an extraneous solution.} \end{aligned}$$

\]
</reasoning>

<answer>
No, because substituting \(x = 4\) into the equation yields \(6 = 0\), which is false. The value \(x = 4\) is an extraneous solution because the principal square root \(\sqrt{2x+1}\) cannot be negative, meaning the equation \(\sqrt{2x+1} = -3\) has no real solutions.
</answer>

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Answer:

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<reasoning>

Substitute the proposed solution

Using the Solving Radical Equations knowledge point
\[

$$\begin{aligned} &\text{Substitute } x = 4 \text{ into the left side of the equation:}\\ &\sqrt{2(4)+1}+3 = \sqrt{8+1}+3 = \sqrt{9}+3 \end{aligned}$$

\]

Evaluate the principal square root

Using the Principal Square Root Definition knowledge point
\[

$$\begin{aligned} &\sqrt{9} = 3\\ &3 + 3 = 6 e 0 \end{aligned}$$

\]

Identify the extraneous solution

Using the Extraneous Solutions knowledge point
\[

$$\begin{aligned} &\sqrt{2x+1} = -3\\ &\text{Since } \sqrt{2x+1} \ge 0 \text{ for all real } x, \text{ there are no real solutions.}\\ &x = 4 \text{ is an extraneous solution.} \end{aligned}$$

\]
</reasoning>

<answer>
No, because substituting \(x = 4\) into the equation yields \(6 = 0\), which is false. The value \(x = 4\) is an extraneous solution because the principal square root \(\sqrt{2x+1}\) cannot be negative, meaning the equation \(\sqrt{2x+1} = -3\) has no real solutions.
</answer>

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