Show me all the way to solve this: Root(x-3)=x-5

The given equation is:

\[
\sqrt{x - 3} = x - 5
\]

To solve this equation, we can follow these steps:

### 1. **Isolate the square root**:
The square root is already isolated on the left side, so we can proceed by squaring both sides of the equation to eliminate the square root.

\[
(\sqrt{x - 3})^2 = (x - 5)^2
\]

This simplifies to:

\[
x - 3 = (x - 5)^2
\]

### 2. **Expand the right side**:
Now, expand the squared binomial \((x - 5)^2\):

\[
x - 3 = (x - 5)(x - 5) = x^2 - 10x + 25
\]

Now the equation is:

\[
x - 3 = x^2 - 10x + 25
\]

### 3. **Move all terms to one side**:
To simplify, let's move all the terms to one side to set the equation to 0:

\[
0 = x^2 - 10x + 25 - x + 3
\]

Simplifying:

\[
0 = x^2 - 11x + 28
\]

### 4. **Solve the quadratic equation**:
Now we have the quadratic equation:

\[
x^2 - 11x + 28 = 0
\]

We can solve this using the **quadratic formula**:

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

For the equation \(x^2 - 11x + 28 = 0\), we have:
- \(a = 1\)
- \(b = -11\)
- \(c = 28\)

Substitute these values into the quadratic formula:

\[
x = \frac{-(-11) \pm \sqrt{(-11)^2 - 4(1)(28)}}{2(1)}
\]

\[
x = \frac{11 \pm \sqrt{121 - 112}}{2}
\]

\[
x = \frac{11 \pm \sqrt{9}}{2}
\]

\[
x = \frac{11 \pm 3}{2}
\]

So, we have two solutions:

\[
x_1 = \frac{11 + 3}{2} = \frac{14}{2} = 7
\]

\[
x_2 = \frac{11 - 3}{2} = \frac{8}{2} = 4
\]

### 5. **Check for extraneous solutions**:
Since we squared both sides in Step 1, we must check whether each solution satisfies the original equation.

- **For \(x = 7\)**:
  \[
  \sqrt{7 - 3} = 7 - 5
  \]
  \[
  \sqrt{4} = 2
  \]
  This is true, so \(x = 7\) is a valid solution.

- **For \(x = 4\)**:
  \[
  \sqrt{4 - 3} = 4 - 5
  \]
  \[
  \sqrt{1} = -1
  \]
  This is false, because \(\sqrt{1} = 1\), not \(-1\). So \(x = 4\) is an extraneous solution.

### 6. **Final Answer**:
The only valid solution is:

\[
x = 7
\]

### Summary of Steps:
1. Square both sides to eliminate the square root.
2. Expand and simplify the resulting quadratic equation.
3. Solve the quadratic equation using the quadratic formula.
4. Check for extraneous solutions.

---

Would you like more details on any of these steps?

### Related Questions:
1. What happens if we use a different method to solve square root equations?
2. How do we identify extraneous solutions in general?
3. Can all quadratic equations be solved using the quadratic formula?
4. What are some alternative methods for solving quadratic equations?
5. How do you solve equations involving cube roots?

### Tip:
Always check for extraneous solutions when you square both sides of an equation, as this process can introduce invalid solutions.

Learn how to solve the equation √(x-3) = x-5 step by step, involving isolating the square root, expanding a quadratic equation, and using the quadratic formula. This problem focuses on algebraic concepts and identifying extraneous solutions, making it suitable for Grades 9-10.

Math Problem Statement

Solution

Ask a new question for Free

By Image

Math Problem Analysis

Mathematical Concepts

Formulas

Theorems

Suitable Grade Level

Related Recommendation