Question

find the real solution of the following radical equation.

(c - \sqrt{14 - 2c} = 7)

(c = ?)

Explanation:

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<reasoning>

Isolate the radical term

\[

$$\begin{aligned} c - 7 = \sqrt{14 - 2c} \end{aligned}$$

\]

Square both sides and solve the quadratic equation

\[

$$\begin{aligned} (c - 7)^2 &= 14 - 2c \\ c^2 - 14c + 49 &= 14 - 2c \\ c^2 - 12c + 35 &= 0 \\ (c - 5)(c - 7) &= 0 \\ c = 5 \quad &\text{or} \quad c = 7 \end{aligned}$$

\]

Check for extraneous solutions

\[

$$\begin{aligned} \text{For } c = 5: \quad 5 - \sqrt{14 - 2(5)} &= 5 - \sqrt{4} = 3 eq 7 \quad (\text{Extraneous}) \\ \text{For } c = 7: \quad 7 - \sqrt{14 - 2(7)} &= 7 - \sqrt{0} = 7 \quad (\text{Valid}) \end{aligned}$$

\]
</reasoning>

<answer>
Find the real solution of the following radical equation.
\(c - \sqrt{14 - 2c} = 7\)
\(c =\) <blank>7</blank>
</answer>

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"Mathematics",
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</post_analysis>

Answer:

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<reasoning>

Isolate the radical term

\[

$$\begin{aligned} c - 7 = \sqrt{14 - 2c} \end{aligned}$$

\]

Square both sides and solve the quadratic equation

\[

$$\begin{aligned} (c - 7)^2 &= 14 - 2c \\ c^2 - 14c + 49 &= 14 - 2c \\ c^2 - 12c + 35 &= 0 \\ (c - 5)(c - 7) &= 0 \\ c = 5 \quad &\text{or} \quad c = 7 \end{aligned}$$

\]

Check for extraneous solutions

\[

$$\begin{aligned} \text{For } c = 5: \quad 5 - \sqrt{14 - 2(5)} &= 5 - \sqrt{4} = 3 eq 7 \quad (\text{Extraneous}) \\ \text{For } c = 7: \quad 7 - \sqrt{14 - 2(7)} &= 7 - \sqrt{0} = 7 \quad (\text{Valid}) \end{aligned}$$

\]
</reasoning>

<answer>
Find the real solution of the following radical equation.
\(c - \sqrt{14 - 2c} = 7\)
\(c =\) <blank>7</blank>
</answer>

<post_analysis>
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"question_type": "Fill-in-the-blank",
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"Mathematics",
"Algebra",
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