To solve the equation $\sqrt{7 - x} = x - 1$, we will square both sides to eliminate the square root. Here's the step-by-step process:

Step 1: Square both sides of the equation

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

Step 2: Expand the square on the right side

$7 - x = x^2 - 2x + 1$

Step 3: Rearrange the equation to set it equal to zero

$0 = x^2 - x - 6$

Step 4: Solve the quadratic equation

We need to factor or use the quadratic formula to solve for $x$.

The quadratic formula is: $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$ For the equation $x^2 - x - 6 = 0$, $a = 1$, $b = -1$, and $c = -6$.

Plugging in the values: $x = \frac{-(-1) \pm \sqrt{(-1)^2 - 4(1)(-6)}}{2(1)}$ $x = \frac{1 \pm \sqrt{1 + 24}}{2}$ $x = \frac{1 \pm \sqrt{25}}{2}$ $x = \frac{1 \pm 5}{2}$ $x = \frac{6}{2} \quad \text{or} \quad x = \frac{-4}{2}$ $x = 3 \quad \text{or} \quad x = -2$

Step 5: Check the solutions in the original equation

1. For $x = 3$: $\sqrt{7 - 3} = 3 - 1$ $\sqrt{4} = 2$ This is true, so $x = 3$ is a valid solution.

2. For $x = -2$: $\sqrt{7 - (-2)} = -2 - 1$ $\sqrt{9} = -3$ This is false because the square root of a number cannot be negative.

Conclusion:

The only valid solution is $\boxed{3}$.

Would you like further details on any part of the solution?

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Here are 5 related questions you might find interesting:

1. What is the general method for solving quadratic equations?
2. How do you solve equations involving square roots and other radicals?
3. What is the importance of checking solutions in the original equation?
4. Can all quadratic equations be factored easily?
5. How do you determine if a quadratic equation has real or complex roots?

Tip: Always check your solutions when dealing with square roots or other nonlinear operations to ensure they satisfy the original equation.