https://file.aiquickdraw.com/mathbot/file/17338391688151owsnytd.jpg

From the uploaded image, I see two mathematical methods for solving quadratic equations: **"By taking square roots"** and **"By completing the square"**. Below, I'll explain both methods in detail based on your notes:

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### **I. Solving by Taking Square Roots**

#### **Key Theorem**:
If \( w^2 = c \), then \( w = \pm \sqrt{c} \).  
This approach works if the quadratic is already in the form \( (expression)^2 = constant \).

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#### **Example**:
Solve \( (3x - 1)^2 = -5 \).  

##### Step 1: Isolate the square term
\[
(3x - 1)^2 = -5
\]

##### Step 2: Take the square root of both sides
\[
|3x - 1| = \sqrt{-5} = \pm i\sqrt{5}
\]

Here, since \( -5 \) is negative, we include \( i \), the imaginary unit.

##### Step 3: Remove the absolute value
\[
3x - 1 = \pm i\sqrt{5}
\]

##### Step 4: Solve for \( x \)
Split into two cases:
\[
3x - 1 = +i\sqrt{5} \quad \text{and} \quad 3x - 1 = -i\sqrt{5}
\]
\[
3x = 1 \pm i\sqrt{5}
\]
\[
x = \frac{1 \pm i\sqrt{5}}{3}
\]

##### Final Answer:
\[
x = \frac{1 + i\sqrt{5}}{3}, \frac{1 - i\sqrt{5}}{3}
\]

---

### **II. Solving by Completing the Square**

This method rewrites the quadratic into a perfect square form \( (x + k)^2 = h \), then solves.

---

#### **Example**:
Solve \( 3x^2 + x = 1 \).

##### Step 1: Divide by the coefficient of \( x^2 \) to make it 1
\[
x^2 + \frac{1}{3}x = \frac{1}{3}
\]

##### Step 2: Add \( \left(\frac{\text{coefficient of } x}{2}\right)^2 \) to both sides  
Here, the coefficient of \( x \) is \( \frac{1}{3} \). Half of it is \( \frac{1}{6} \), and squaring gives \( \left(\frac{1}{6}\right)^2 = \frac{1}{36} \). Add this to both sides:
\[
x^2 + \frac{1}{3}x + \frac{1}{36} = \frac{1}{3} + \frac{1}{36}
\]

##### Step 3: Simplify the equation
The left-hand side becomes a perfect square:
\[
\left(x + \frac{1}{6}\right)^2 = \frac{12}{36} + \frac{1}{36} = \frac{13}{36}
\]

##### Step 4: Solve by taking square roots
\[
x + \frac{1}{6} = \pm \sqrt{\frac{13}{36}}
\]
\[
x + \frac{1}{6} = \pm \frac{\sqrt{13}}{6}
\]

##### Step 5: Solve for \( x \)
\[
x = -\frac{1}{6} \pm \frac{\sqrt{13}}{6}
\]
\[
x = \frac{-1 \pm \sqrt{13}}{6}
\]

##### Final Answer:
\[
x = \frac{-1 + \sqrt{13}}{6}, \frac{-1 - \sqrt{13}}{6}
\]

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### Summary

- **Method I**: Directly solves for \( x \) by isolating and taking square roots. Works well with expressions like \( (expression)^2 = c \).
- **Method II**: Transforms the quadratic into a perfect square and solves. Useful for general quadratics.

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Do you want further details or clarification?  
Here are some related questions for practice:

1. Solve \( (x - 3)^2 = 16 \) using the square root method.
2. Solve \( 2x^2 + 8x + 6 = 0 \) by completing the square.
3. What is the difference between real and imaginary roots in a quadratic equation?
4. How does completing the square relate to the quadratic formula?
5. Can all quadratics be solved using the square root method? Why or why not?

**Tip**: Always simplify coefficients and constants first to make calculations easier.

Understand how to solve quadratic equations using two methods: taking square roots and completing the square. This tutorial explains each step with examples, including cases involving imaginary numbers. Suitable for high school students.

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