Solve the radical equation. Be sure to check all solutions to eliminate extraneous solutions. (Enter your answers as a comma-separated list. If an answer does not exist, enter DNE.)

Square root of (x − 1) = x − 7

To solve the radical equation \( \sqrt{x - 1} = x - 7 \), follow these steps:

### Step 1: Isolate the square root.
The equation is already in the form where the square root is isolated on the left-hand side.

\[
\sqrt{x - 1} = x - 7
\]

### Step 2: Square both sides to eliminate the square root.
Squaring both sides of the equation gives:

\[
(\sqrt{x - 1})^2 = (x - 7)^2
\]

This simplifies to:

\[
x - 1 = (x - 7)^2
\]

### Step 3: Expand the square on the right side.
\[
x - 1 = (x - 7)(x - 7) = x^2 - 14x + 49
\]

Now the equation becomes:

\[
x - 1 = x^2 - 14x + 49
\]

### Step 4: Rearrange the equation into a standard quadratic form.
Move all terms to one side of the equation:

\[
0 = x^2 - 14x + 49 - x + 1
\]

Simplify:

\[
0 = x^2 - 15x + 50
\]

### Step 5: Solve the quadratic equation.
We can solve \( x^2 - 15x + 50 = 0 \) using the quadratic formula. The quadratic formula is:

\[
x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}
\]

For the equation \( x^2 - 15x + 50 = 0 \), the coefficients are \( a = 1 \), \( b = -15 \), and \( c = 50 \). Substituting into the formula:

\[
x = \frac{-(-15) \pm \sqrt{(-15)^2 - 4(1)(50)}}{2(1)}
\]
\[
x = \frac{15 \pm \sqrt{225 - 200}}{2}
\]
\[
x = \frac{15 \pm \sqrt{25}}{2}
\]
\[
x = \frac{15 \pm 5}{2}
\]

So the two solutions are:

\[
x_1 = \frac{15 + 5}{2} = \frac{20}{2} = 10
\]
\[
x_2 = \frac{15 - 5}{2} = \frac{10}{2} = 5
\]

### Step 6: Check for extraneous solutions.
We must check both solutions in the original equation \( \sqrt{x - 1} = x - 7 \).

- For \( x = 10 \):

\[
\sqrt{10 - 1} = \sqrt{9} = 3
\]
\[
10 - 7 = 3
\]

This is a valid solution.

- For \( x = 5 \):

\[
\sqrt{5 - 1} = \sqrt{4} = 2
\]
\[
5 - 7 = -2
\]

This is not valid because the square root function only gives non-negative values, but we got a negative number on the right side. So, \( x = 5 \) is an extraneous solution.

### Final Answer:
The only solution is:

\[
x = 10
\]

Would you like any additional explanation or details?

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Here are 5 related questions you might want to explore:
1. How do you solve equations with two square roots?
2. How do you check for extraneous solutions in general?
3. What are other methods to solve radical equations?
4. How would you solve a cube root equation?
5. How does graphing help visualize solutions to radical equations?

**Tip:** Always check for extraneous solutions when solving radical equations by plugging your solutions back into the original equation.

Solve the radical equation. Be sure to check all solutions to eliminate extraneous solutions. (Enter your answers as a comma-separated list. If an answer does not exist, enter DNE.) Square root of (x − 1) = x − 7

Learn how to solve the radical equation √(x - 1) = x - 7 with a detailed step-by-step solution. This problem covers key algebraic concepts, including radical equations, quadratic formula, and checking for extraneous solutions. Suitable for students in Grades 9-11.

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