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:

---

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 $(expression)^2 = constant$.

---

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:

$x = \frac{-1 + \sqrt{13}}{6}, \frac{-1 - \sqrt{13}}{6}$

---

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.

---

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.