Question

solve: \\(\sqrt3{x^2 - 8} = 2\\)

• \\(x = -4\\)
• \\(x = 4\\)
• \\(x = -4\\) or \\(x = 4\\)
• no real solution

solve: \\(s = 4 + \sqrt{s + 2}\\)

• \\(s = 2\\)
• \\(s = 7\\)
• \\(s = 2\\) or \\(s = 7\\)
• no real solution

Explanation:

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<reasoning>

Solve the cube root equation

Using the Cube Root Equations knowledge point
\[

$$\begin{aligned} \sqrt[3]{x^2 - 8} &= 2 \\ x^2 - 8 &= 2^3 \\ x^2 - 8 &= 8 \\ x^2 &= 16 \\ x &= \pm 4 \end{aligned}$$

\]

Solve the square root equation

Using the Solving Radical Equations knowledge point
\[

$$\begin{aligned} s &= 4 + \sqrt{s + 2} \\ s - 4 &= \sqrt{s + 2} \\ (s - 4)^2 &= s + 2 \\ s^2 - 8s + 16 &= s + 2 \\ s^2 - 9s + 14 &= 0 \\ (s - 7)(s - 2) &= 0 \\ s = 7 \quad &\text{or} \quad s = 2 \end{aligned}$$

\]

Check for extraneous solutions

Using the Solving Radical Equations knowledge point
\[

$$\begin{aligned} &\text{For } s = 7: \quad 7 \stackrel{?}{=} 4 + \sqrt{7 + 2} \implies 7 = 4 + 3 \quad (\text{True}) \\ &\text{For } s = 2: \quad 2 \stackrel{?}{=} 4 + \sqrt{2 + 2} \implies 2 = 4 + 2 \quad (\text{False}) \end{aligned}$$

\]
Thus, \(s = 7\) is the only real solution.
</reasoning>

<answer>

Question 1

<mcq-option>(A) \(x = -4\)</mcq-option>
<mcq-option>(B) \(x = 4\)</mcq-option>
<mcq-correct>(C) \(x = -4 \text{ or } x = 4\)</mcq-correct>
<mcq-option>(D) no real solution</mcq-option>

Question 2

<mcq-option>(A) \(s = 2\)</mcq-option>
<mcq-correct>(B) \(s = 7\)</mcq-correct>
<mcq-option>(C) \(s = 2 \text{ or } s = 7\)</mcq-option>
<mcq-option>(D) no real solution</mcq-option>
</answer>

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"question_type": "Multiple Choice",
"knowledge_point": [
"Mathematics",
"Algebra",
"Solving Radical Equations"
]
}
</post_analysis>

Answer:

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<reasoning>

Solve the cube root equation

Using the Cube Root Equations knowledge point
\[

$$\begin{aligned} \sqrt[3]{x^2 - 8} &= 2 \\ x^2 - 8 &= 2^3 \\ x^2 - 8 &= 8 \\ x^2 &= 16 \\ x &= \pm 4 \end{aligned}$$

\]

Solve the square root equation

Using the Solving Radical Equations knowledge point
\[

$$\begin{aligned} s &= 4 + \sqrt{s + 2} \\ s - 4 &= \sqrt{s + 2} \\ (s - 4)^2 &= s + 2 \\ s^2 - 8s + 16 &= s + 2 \\ s^2 - 9s + 14 &= 0 \\ (s - 7)(s - 2) &= 0 \\ s = 7 \quad &\text{or} \quad s = 2 \end{aligned}$$

\]

Check for extraneous solutions

Using the Solving Radical Equations knowledge point
\[

$$\begin{aligned} &\text{For } s = 7: \quad 7 \stackrel{?}{=} 4 + \sqrt{7 + 2} \implies 7 = 4 + 3 \quad (\text{True}) \\ &\text{For } s = 2: \quad 2 \stackrel{?}{=} 4 + \sqrt{2 + 2} \implies 2 = 4 + 2 \quad (\text{False}) \end{aligned}$$

\]
Thus, \(s = 7\) is the only real solution.
</reasoning>

<answer>

Question 1

<mcq-option>(A) \(x = -4\)</mcq-option>
<mcq-option>(B) \(x = 4\)</mcq-option>
<mcq-correct>(C) \(x = -4 \text{ or } x = 4\)</mcq-correct>
<mcq-option>(D) no real solution</mcq-option>

Question 2

<mcq-option>(A) \(s = 2\)</mcq-option>
<mcq-correct>(B) \(s = 7\)</mcq-correct>
<mcq-option>(C) \(s = 2 \text{ or } s = 7\)</mcq-option>
<mcq-option>(D) no real solution</mcq-option>
</answer>

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"knowledge_point": [
"Mathematics",
"Algebra",
"Solving Radical Equations"
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