We will solve the equation:

$\sqrt{x-5} + x = 7$

Step 1: Isolate the square root term

Move $x$ to the other side of the equation: $\sqrt{x-5} = 7 - x$

Step 2: Square both sides

Squaring both sides to eliminate the square root: $(\sqrt{x-5})^2 = (7 - x)^2$ $x-5 = 49 - 14x + x^2$

Step 3: Rearrange into a standard quadratic form

Bring all terms to one side: $x^2 - 15x + 54 = 0$

Step 4: Factorize the quadratic

We factorize $x^2 - 15x + 54$: $(x - 9)(x - 6) = 0$

Step 5: Solve for $x$

The solutions are: $x = 9 \quad \text{or} \quad x = 6$

Step 6: Verify the solutions

We check each solution in the original equation to ensure they work.

For $x = 9$:

$\sqrt{9-5} + 9 = \sqrt{4} + 9 = 2 + 9 = 7$ This is not true, so $x = 9$ is extraneous.

For $x = 6$:

$\sqrt{6-5} + 6 = \sqrt{1} + 6 = 1 + 6 = 7$ This is true, so $x = 6$ is valid.

Final Solution:

$x = 6$

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Would you like additional explanations or a deeper dive into any step?

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Related Questions:

1. How do you identify extraneous solutions in square root equations?
2. What are the common steps for solving quadratic equations?
3. Can you explain how to check solutions in equations involving square roots?
4. What does squaring both sides of an equation imply about potential errors?
5. How can factoring techniques be applied to more complex quadratic equations?

Tip:

Always verify your solutions when solving equations involving square roots, as squaring both sides can introduce extraneous solutions.