Question

study the steps shown to solve the given equation.
\\(\sqrt{30 - 2x} = x - 3\\)
\\(30 - 2x = x^2 - 6x + 9\\)
\\(0 = x^2 - 4x - 21\\)
\\(0 = (x + 3)(x - 7)\\)
based on the above work, possible solutions of the equation are \\(-3\\) and \\(7\\).
check the answers in the original equation. the extraneous solution is \\(x = \\)

Explanation:

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<reasoning>

Identify the potential solutions

Using the Solving Radical Equations knowledge point
\[

$$\begin{aligned} &\sqrt{30-2x} = x-3 \\ &30-2x = (x-3)^2 = x^2-6x+9 \\ &x^2-4x-21 = 0 \\ &(x+3)(x-7) = 0 \\ &x = -3 \quad \text{or} \quad x = 7 \end{aligned}$$

\]

Verify the potential solutions

Using the Extraneous Solutions knowledge point
\[

$$\begin{aligned} &\text{For } x = -3:\\ &\text{LHS} = \sqrt{30-2(-3)} = \sqrt{36} = 6 \\ &\text{RHS} = -3-3 = -6 \\ &\text{LHS} e \text{RHS} \implies x = -3 \text{ is extraneous.} \\ \\ &\text{For } x = 7:\\ &\text{LHS} = \sqrt{30-2(7)} = \sqrt{16} = 4 \\ &\text{RHS} = 7-3 = 4 \\ &\text{LHS} = \text{RHS} \implies x = 7 \text{ is a valid solution.} \end{aligned}$$

\]
</reasoning>

<answer>
Check the answers in the original equation. The extraneous solution is \(x =\) <blank>-3</blank>
</answer>

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Answer:

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<reasoning>

Identify the potential solutions

Using the Solving Radical Equations knowledge point
\[

$$\begin{aligned} &\sqrt{30-2x} = x-3 \\ &30-2x = (x-3)^2 = x^2-6x+9 \\ &x^2-4x-21 = 0 \\ &(x+3)(x-7) = 0 \\ &x = -3 \quad \text{or} \quad x = 7 \end{aligned}$$

\]

Verify the potential solutions

Using the Extraneous Solutions knowledge point
\[

$$\begin{aligned} &\text{For } x = -3:\\ &\text{LHS} = \sqrt{30-2(-3)} = \sqrt{36} = 6 \\ &\text{RHS} = -3-3 = -6 \\ &\text{LHS} e \text{RHS} \implies x = -3 \text{ is extraneous.} \\ \\ &\text{For } x = 7:\\ &\text{LHS} = \sqrt{30-2(7)} = \sqrt{16} = 4 \\ &\text{RHS} = 7-3 = 4 \\ &\text{LHS} = \text{RHS} \implies x = 7 \text{ is a valid solution.} \end{aligned}$$

\]
</reasoning>

<answer>
Check the answers in the original equation. The extraneous solution is \(x =\) <blank>-3</blank>
</answer>

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