\[\sqrt{6x-6}=x-1\]
'
What is the product of all solutions to the given equation?

To solve the equation \(\sqrt{6x - 6} = x - 1\), we will follow these steps:

### Step 1: Square both sides
Square both sides of the equation to eliminate the square root:

\[
(\sqrt{6x - 6})^2 = (x - 1)^2
\]

This simplifies to:

\[
6x - 6 = (x - 1)^2
\]

### Step 2: Expand the quadratic on the right side
Expand \((x - 1)^2\):

\[
6x - 6 = x^2 - 2x + 1
\]

### Step 3: Rearrange the equation
Bring all terms to one side to form a quadratic equation:

\[
x^2 - 2x - 6x + 1 + 6 = 0
\]

Simplify this:

\[
x^2 - 8x + 7 = 0
\]

### Step 4: Solve the quadratic equation
Now, solve the quadratic equation \(x^2 - 8x + 7 = 0\). The solutions can be found using the quadratic formula:

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

For our equation \(x^2 - 8x + 7 = 0\), \(a = 1\), \(b = -8\), and \(c = 7\). Plugging in these values:

\[
x = \frac{-(-8) \pm \sqrt{(-8)^2 - 4(1)(7)}}{2(1)}
\]

\[
x = \frac{8 \pm \sqrt{64 - 28}}{2}
\]

\[
x = \frac{8 \pm \sqrt{36}}{2}
\]

\[
x = \frac{8 \pm 6}{2}
\]

So, the two possible solutions are:

\[
x = \frac{8 + 6}{2} = 7 \quad \text{and} \quad x = \frac{8 - 6}{2} = 1
\]

### Step 5: Verify the solutions
Substitute \(x = 7\) and \(x = 1\) back into the original equation to check for extraneous solutions:

- For \(x = 7\):

\[
\sqrt{6(7) - 6} = \sqrt{42 - 6} = \sqrt{36} = 6
\]

\[
x - 1 = 7 - 1 = 6
\]

Both sides are equal, so \(x = 7\) is a valid solution.

- For \(x = 1\):

\[
\sqrt{6(1) - 6} = \sqrt{6 - 6} = \sqrt{0} = 0
\]

\[
x - 1 = 1 - 1 = 0
\]

Both sides are equal, so \(x = 1\) is also a valid solution.

### Step 6: Calculate the product of all solutions
The solutions are \(x = 7\) and \(x = 1\). The product of all solutions is:

\[
7 \times 1 = 7
\]

Thus, the product of all solutions is \(\boxed{7}\).

Would you like further details on any step or have any questions?

Here are some related questions you might find interesting:
1. What happens if the quadratic equation had no real solutions?
2. How do you solve a quadratic equation by factoring?
3. What is the discriminant and how does it determine the nature of the roots?
4. Can the product of roots be found directly from the quadratic equation's coefficients?
5. What are extraneous solutions, and why do they appear in equations involving square roots?

**Tip:** Always verify your solutions by substituting them back into the original equation, especially when dealing with square roots or other operations that can introduce extraneous solutions.

Learn how to solve the equation √(6x - 6) = x - 1 step-by-step with detailed solutions. This problem involves equations with square roots and quadratic elements, suitable for advanced high school students.

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