We are given the equation:

$\sqrt{2x - 3} = x - 3$

To solve this equation, let's follow these steps:

Step 1: Square both sides

First, square both sides of the equation to eliminate the square root:

$(\sqrt{2x - 3})^2 = (x - 3)^2$

This simplifies to:

$2x - 3 = (x - 3)^2$

Step 2: Expand the right side

Now, expand $(x - 3)^2$:

$2x - 3 = x^2 - 6x + 9$

Step 3: Rearrange the equation

Bring all terms to one side to set the equation to 0:

$0 = x^2 - 6x + 9 - 2x + 3$

Simplify the terms:

$0 = x^2 - 8x + 12$

Step 4: Solve the quadratic equation

The equation now is:

$x^2 - 8x + 12 = 0$

We can solve this using the quadratic formula:

$x = \frac{-(-8) \pm \sqrt{(-8)^2 - 4(1)(12)}}{2(1)}$

Simplifying:

$x = \frac{8 \pm \sqrt{64 - 48}}{2}$

$x = \frac{8 \pm \sqrt{16}}{2}$

$x = \frac{8 \pm 4}{2}$

Thus, we have two possible solutions:

$x = \frac{8 + 4}{2} = 6 \quad \text{or} \quad x = \frac{8 - 4}{2} = 2$

Step 5: Check for extraneous solutions

We need to check these solutions in the original equation because squaring both sides may introduce extraneous solutions.

• For $x = 6$:

$\sqrt{2(6) - 3} = 6 - 3 \quad \Rightarrow \quad \sqrt{12 - 3} = 3 \quad \Rightarrow \quad \sqrt{9} = 3 \quad \text{(True)}$

• For $x = 2$:

\sqrt{2(2) - 3} = 2 - 3 \quad \Rightarrow \quad \sqrt{4 - 3} = -1 \quad \Rightarrow \quad \sqrt{1} = -1 \quad \text{(False, because \sqrt{1} is always 1, not -1)}

Thus, $x = 2$ is an extraneous solution.

Final Solution:

The only valid solution is:

$x = 6$

Do you want more details or have any questions?

Related Questions:

1. How do you solve equations involving square roots and quadratic forms?
2. What are extraneous solutions, and why do they occur?
3. Can all quadratic equations be solved by factoring instead of the quadratic formula?
4. How do you check if a solution to a radical equation is valid?
5. What are the steps for solving equations with multiple radicals?

Tip:

Always check for extraneous solutions when solving equations involving square roots or squaring both sides!