Question

find the real solution of the following radical equation.

\\b - 10 = \sqrt{20 - 2b}\\

\\b = ?\\

Explanation:

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<reasoning>

Square both sides of the equation

\[

$$\begin{aligned} (b - 10)^2 &= (\sqrt{20 - 2b})^2 \\ b^2 - 20b + 100 &= 20 - 2b \end{aligned}$$

\]

Rearrange into standard quadratic form

\[

$$\begin{aligned} b^2 - 18b + 80 = 0 \end{aligned}$$

\]

Factor and solve for b, then check for extraneous solutions

\[

$$\begin{aligned} (b - 8)(b - 10) &= 0 \\ b = 8 \quad &\text{or} \quad b = 10 \end{aligned}$$

\]
Checking \(b = 8\):
\[

$$\begin{aligned} 8 - 10 &\stackrel{?}{=} \sqrt{20 - 2(8)} \\ -2 & e \sqrt{4} \quad (\text{extraneous}) \end{aligned}$$

\]
Checking \(b = 10\):
\[

$$\begin{aligned} 10 - 10 &\stackrel{?}{=} \sqrt{20 - 2(10)} \\ 0 &= \sqrt{0} \quad (\text{valid}) \end{aligned}$$

\]
</reasoning>

<answer>
Find the real solution of the following radical equation.

\(b - 10 = \sqrt{20 - 2b}\)

\(b =\) <blank>10</blank>
</answer>

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Answer:

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<reasoning>

Square both sides of the equation

\[

$$\begin{aligned} (b - 10)^2 &= (\sqrt{20 - 2b})^2 \\ b^2 - 20b + 100 &= 20 - 2b \end{aligned}$$

\]

Rearrange into standard quadratic form

\[

$$\begin{aligned} b^2 - 18b + 80 = 0 \end{aligned}$$

\]

Factor and solve for b, then check for extraneous solutions

\[

$$\begin{aligned} (b - 8)(b - 10) &= 0 \\ b = 8 \quad &\text{or} \quad b = 10 \end{aligned}$$

\]
Checking \(b = 8\):
\[

$$\begin{aligned} 8 - 10 &\stackrel{?}{=} \sqrt{20 - 2(8)} \\ -2 & e \sqrt{4} \quad (\text{extraneous}) \end{aligned}$$

\]
Checking \(b = 10\):
\[

$$\begin{aligned} 10 - 10 &\stackrel{?}{=} \sqrt{20 - 2(10)} \\ 0 &= \sqrt{0} \quad (\text{valid}) \end{aligned}$$

\]
</reasoning>

<answer>
Find the real solution of the following radical equation.

\(b - 10 = \sqrt{20 - 2b}\)

\(b =\) <blank>10</blank>
</answer>

<post_analysis>
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