Question

find the real solution of the following radical equation.

\\\sqrt{y-1} = y-3\\

\\y = ?\\

Explanation:

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<reasoning>

Square both sides of the equation

Using the Solving Radical Equations knowledge point
\[

$$\begin{aligned} \sqrt{y-1} &= y-3 \\ (\sqrt{y-1})^2 &= (y-3)^2 \\ y-1 &= y^2 - 6y + 9 \end{aligned}$$

\]

Rearrange into standard quadratic form

Using the Solving Radical Equations knowledge point
\[

$$\begin{aligned} y^2 - 6y - y + 9 + 1 &= 0 \\ y^2 - 7y + 10 &= 0 \end{aligned}$$

\]

Solve the quadratic equation

Using the Solving Radical Equations knowledge point
\[

$$\begin{aligned} (y-2)(y-5) &= 0 \\ y = 2 \quad &\text{or} \quad y = 5 \end{aligned}$$

\]

Check for extraneous solutions

We must verify both potential solutions in the original equation \(\sqrt{y-1} = y-3\).

For \(y = 2\):
\[

$$\begin{aligned} \text{LHS} &= \sqrt{2-1} = \sqrt{1} = 1 \\ \text{RHS} &= 2-3 = -1 \\ 1 & e -1 \quad (\text{Extraneous}) \end{aligned}$$

\]

For \(y = 5\):
\[

$$\begin{aligned} \text{LHS} &= \sqrt{5-1} = \sqrt{4} = 2 \\ \text{RHS} &= 5-3 = 2 \\ 2 &= 2 \quad (\text{Valid}) \end{aligned}$$

\]
</reasoning>

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

\[\sqrt{y-1} = y-3\]

\(y =\) <blank>5</blank>
</answer>

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Answer:

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<reasoning>

Square both sides of the equation

Using the Solving Radical Equations knowledge point
\[

$$\begin{aligned} \sqrt{y-1} &= y-3 \\ (\sqrt{y-1})^2 &= (y-3)^2 \\ y-1 &= y^2 - 6y + 9 \end{aligned}$$

\]

Rearrange into standard quadratic form

Using the Solving Radical Equations knowledge point
\[

$$\begin{aligned} y^2 - 6y - y + 9 + 1 &= 0 \\ y^2 - 7y + 10 &= 0 \end{aligned}$$

\]

Solve the quadratic equation

Using the Solving Radical Equations knowledge point
\[

$$\begin{aligned} (y-2)(y-5) &= 0 \\ y = 2 \quad &\text{or} \quad y = 5 \end{aligned}$$

\]

Check for extraneous solutions

We must verify both potential solutions in the original equation \(\sqrt{y-1} = y-3\).

For \(y = 2\):
\[

$$\begin{aligned} \text{LHS} &= \sqrt{2-1} = \sqrt{1} = 1 \\ \text{RHS} &= 2-3 = -1 \\ 1 & e -1 \quad (\text{Extraneous}) \end{aligned}$$

\]

For \(y = 5\):
\[

$$\begin{aligned} \text{LHS} &= \sqrt{5-1} = \sqrt{4} = 2 \\ \text{RHS} &= 5-3 = 2 \\ 2 &= 2 \quad (\text{Valid}) \end{aligned}$$

\]
</reasoning>

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

\[\sqrt{y-1} = y-3\]

\(y =\) <blank>5</blank>
</answer>

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