Question

Solve the equation below. Be sure to check for extraneous solutions. x−1−−−−√+4 = x−3

Answer 1

To solve the equation \( \sqrt{x-1} + 4 = x - 3 \), follow these steps:

1. **Isolate the square root:**
\[
\sqrt{x-1} = x - 3 - 4
\]
\[
\sqrt{x-1} = x - 7
\]

2. **Square both sides to eliminate the square root:**
\[
(\sqrt{x-1})^2 = (x - 7)^2
\]
\[
x - 1 = (x - 7)(x - 7)
\]
\[
x - 1 = x^2 - 14x + 49
\]

3. **Rearrange the equation:**
\[
0 = x^2 - 14x + 49 - x + 1
\]
\[
0 = x^2 - 15x + 50
\]

4. **Factor or use the quadratic formula to solve for \( x \):**
The quadratic is \( x^2 - 15x + 50 = 0 \). We can factor this as:
\[
(x - 10)(x - 5) = 0
\]
which gives us:
\[
x = 10 \quad \text{or} \quad x = 5
\]

5. **Check for extraneous solutions:**

For \( x = 10 \):
\[
\sqrt{10-1} + 4 = 3 + 4 = 7
\]
\[
10 - 3 = 7
\]
Thus, \( x = 10 \) is a valid solution.

For \( x = 5 \):
\[
\sqrt{5-1} + 4 = 2 + 4 = 6
\]
\[
5 - 3 = 2
\]
Thus, \( x = 5 \) does not satisfy the original equation.

6. **Final solution:**
The only solution to the equation is
\[
\boxed{10}
\]