Question

find the real solution of the following radical equation.

\\24\sqrt{2b} - 19 = 29\\

\\b = ?\\

Explanation:

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<reasoning>

Isolate the radical term

\[

$$\begin{aligned} 24\sqrt{2b} - 19 &= 29 \\ 24\sqrt{2b} &= 29 + 19 \\ 24\sqrt{2b} &= 48 \end{aligned}$$

\]

Divide by the coefficient

\[

$$\begin{aligned} \sqrt{2b} &= \frac{48}{24} \\ \sqrt{2b} &= 2 \end{aligned}$$

\]

Square both sides and solve for b

\[

$$\begin{aligned} (\sqrt{2b})^2 &= 2^2 \\ 2b &= 4 \\ b &= 2 \end{aligned}$$

\]
</reasoning>

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

\(24\sqrt{2b} - 19 = 29\)

\(b =\) <blank>2</blank>
</answer>

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"question_type": "Fill-in-the-blank",
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"Mathematics",
"Algebra",
"Solving Radical Equations"
]
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Answer:

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</pre_analysis>

<reasoning>

Isolate the radical term

\[

$$\begin{aligned} 24\sqrt{2b} - 19 &= 29 \\ 24\sqrt{2b} &= 29 + 19 \\ 24\sqrt{2b} &= 48 \end{aligned}$$

\]

Divide by the coefficient

\[

$$\begin{aligned} \sqrt{2b} &= \frac{48}{24} \\ \sqrt{2b} &= 2 \end{aligned}$$

\]

Square both sides and solve for b

\[

$$\begin{aligned} (\sqrt{2b})^2 &= 2^2 \\ 2b &= 4 \\ b &= 2 \end{aligned}$$

\]
</reasoning>

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

\(24\sqrt{2b} - 19 = 29\)

\(b =\) <blank>2</blank>
</answer>

<post_analysis>
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"Mathematics",
"Algebra",
"Solving Radical Equations"
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