Question

a student stated that there are two solutions to the equation \\(\sqrt{3x + 7} = x - 1\\).

they are \\(x = -1\\) and \\(x = 6\\).

do you agree with the student? explain why or why not.

• i do not agree with the student because there is no real solution.
• checking the solutions in the original equation verifies \\(x = 6\\); however, \\(x = -1\\) is extraneous.
• checking the solutions in the original equation shows that 6 is an extraneous solution.
• i agree with the student because both \\(x = -1\\) and \\(x = 6\\) satisfy the original equation.

Explanation:

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<reasoning>

Solve the radical equation

Using the Solving Radical Equations knowledge point
\[

$$\begin{aligned} \sqrt{3x + 7} &= x - 1 \\ 3x + 7 &= (x - 1)^2 \\ 3x + 7 &= x^2 - 2x + 1 \\ x^2 - 5x - 6 &= 0 \\ (x - 6)(x + 1) &= 0 \\ x = 6 \quad &\text{or} \quad x = -1 \end{aligned}$$

\]

Verify the solutions

Using the Extraneous Solutions and Equation Verification knowledge points
\[

$$\begin{aligned} \text{For } x = 6: \quad &\sqrt{3(6) + 7} = \sqrt{25} = 5, \quad 6 - 1 = 5 \quad (\text{Valid}) \\ \text{For } x = -1: \quad &\sqrt{3(-1) + 7} = \sqrt{4} = 2, \quad -1 - 1 = -2 \quad (2 eq -2, \text{ Extraneous}) \end{aligned}$$

\]
</reasoning>

<answer>
<mcq-option>I do not agree with the student because there is no real solution.</mcq-option>
<mcq-correct>Checking the solutions in the original equation verifies \(x = 6\); however, \(x = -1\) is extraneous.</mcq-correct>
<mcq-option>Checking the solutions in the original equation shows that 6 is an extraneous solution.</mcq-option>
<mcq-option>I agree with the student because both \(x = -1\) and \(x = 6\) satisfy the original equation.</mcq-option>
</answer>

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Answer:

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<reasoning>

Solve the radical equation

Using the Solving Radical Equations knowledge point
\[

$$\begin{aligned} \sqrt{3x + 7} &= x - 1 \\ 3x + 7 &= (x - 1)^2 \\ 3x + 7 &= x^2 - 2x + 1 \\ x^2 - 5x - 6 &= 0 \\ (x - 6)(x + 1) &= 0 \\ x = 6 \quad &\text{or} \quad x = -1 \end{aligned}$$

\]

Verify the solutions

Using the Extraneous Solutions and Equation Verification knowledge points
\[

$$\begin{aligned} \text{For } x = 6: \quad &\sqrt{3(6) + 7} = \sqrt{25} = 5, \quad 6 - 1 = 5 \quad (\text{Valid}) \\ \text{For } x = -1: \quad &\sqrt{3(-1) + 7} = \sqrt{4} = 2, \quad -1 - 1 = -2 \quad (2 eq -2, \text{ Extraneous}) \end{aligned}$$

\]
</reasoning>

<answer>
<mcq-option>I do not agree with the student because there is no real solution.</mcq-option>
<mcq-correct>Checking the solutions in the original equation verifies \(x = 6\); however, \(x = -1\) is extraneous.</mcq-correct>
<mcq-option>Checking the solutions in the original equation shows that 6 is an extraneous solution.</mcq-option>
<mcq-option>I agree with the student because both \(x = -1\) and \(x = 6\) satisfy the original equation.</mcq-option>
</answer>

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