Question

find the real solutions of the following radical equation.

\\d = \sqrt{13d - 36}\\

\\d = ?, \\

enter the smallest solution first.

Explanation:

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<reasoning>

Square both sides of the equation

\[

$$\begin{aligned} d &= \sqrt{13d - 36} \\ d^2 &= 13d - 36 \end{aligned}$$

\]

Solve the quadratic equation

\[

$$\begin{aligned} d^2 - 13d + 36 &= 0 \\ (d - 4)(d - 9) &= 0 \\ d = 4 \quad &\text{or} \quad d = 9 \end{aligned}$$

\]

Verify solutions in the original equation

\[

$$\begin{aligned} \text{For } d = 4: \quad 4 &= \sqrt{13(4) - 36} = \sqrt{52 - 36} = \sqrt{16} = 4 \quad (\text{True}) \\ \text{For } d = 9: \quad 9 &= \sqrt{13(9) - 36} = \sqrt{117 - 36} = \sqrt{81} = 9 \quad (\text{True}) \end{aligned}$$

\]
</reasoning>

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

\(d = \sqrt{13d - 36}\)

\(d =\) <blank>4</blank>, <blank>9</blank>

Enter the smallest solution first.
</answer>

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

Answer:

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<reasoning>

Square both sides of the equation

\[

$$\begin{aligned} d &= \sqrt{13d - 36} \\ d^2 &= 13d - 36 \end{aligned}$$

\]

Solve the quadratic equation

\[

$$\begin{aligned} d^2 - 13d + 36 &= 0 \\ (d - 4)(d - 9) &= 0 \\ d = 4 \quad &\text{or} \quad d = 9 \end{aligned}$$

\]

Verify solutions in the original equation

\[

$$\begin{aligned} \text{For } d = 4: \quad 4 &= \sqrt{13(4) - 36} = \sqrt{52 - 36} = \sqrt{16} = 4 \quad (\text{True}) \\ \text{For } d = 9: \quad 9 &= \sqrt{13(9) - 36} = \sqrt{117 - 36} = \sqrt{81} = 9 \quad (\text{True}) \end{aligned}$$

\]
</reasoning>

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

\(d = \sqrt{13d - 36}\)

\(d =\) <blank>4</blank>, <blank>9</blank>

Enter the smallest solution first.
</answer>

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