https://file.aiquickdraw.com/mathbot/file/17300708001901tf0p6ee.jpg

Let's solve each part of the problem provided in the activity.

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### 1. Making a Complete Square

To make a quadratic expression a perfect square, we add the square of half the coefficient of the \( x \)-term.

#### (a) \( 0 = x^2 - 8x + ? \)

1. Take half of the coefficient of \( x \): \( \frac{-8}{2} = -4 \).
2. Square this result: \( (-4)^2 = 16 \).

So, the term to add is **16**.

#### (b) \( y = x^2 + 9x + ? \)

1. Take half of the coefficient of \( x \): \( \frac{9}{2} = 4.5 \).
2. Square this result: \( (4.5)^2 = 20.25 \).

So, the term to add is **20.25**.

#### (c) \( y = x^2 - \frac{b}{a}x + ? \)

1. Take half of the coefficient of \( x \): \( \frac{-\frac{b}{a}}{2} = -\frac{b}{2a} \).
2. Square this result: \( \left(-\frac{b}{2a}\right)^2 = \frac{b^2}{4a^2} \).

So, the term to add is **\( \frac{b^2}{4a^2} \)**.

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### 2. Solving by Completing the Square

In this section, we need to solve each equation by completing the square.

#### (a) \( -3x^2 + 5x + 4 = 0 \)

1. First, divide by \(-3\) to make the coefficient of \( x^2 \) equal to \( 1 \):
   \[
   x^2 - \frac{5}{3}x - \frac{4}{3} = 0
   \]

2. Move the constant term to the other side:
   \[
   x^2 - \frac{5}{3}x = \frac{4}{3}
   \]

3. Complete the square:
   - Half of \(-\frac{5}{3}\) is \(-\frac{5}{6}\).
   - Square it to get \(\frac{25}{36}\).
   - Add \(\frac{25}{36}\) to both sides:
     \[
     x^2 - \frac{5}{3}x + \frac{25}{36} = \frac{4}{3} + \frac{25}{36}
     \]

4. Simplify the right side:
   \[
   \frac{4}{3} = \frac{48}{36} \Rightarrow \frac{48}{36} + \frac{25}{36} = \frac{73}{36}
   \]

5. Write the left side as a square:
   \[
   \left(x - \frac{5}{6}\right)^2 = \frac{73}{36}
   \]

6. Solve for \( x \) by taking the square root of both sides:
   \[
   x - \frac{5}{6} = \pm \sqrt{\frac{73}{36}} = \pm \frac{\sqrt{73}}{6}
   \]

7. Therefore,
   \[
   x = \frac{5 \pm \sqrt{73}}{6}
   \]

#### (b) \( ax^2 + bx + c = 0 \)

To solve this general form by completing the square:

1. Divide by \( a \) to make the coefficient of \( x^2 \) equal to \( 1 \):
   \[
   x^2 + \frac{b}{a}x + \frac{c}{a} = 0
   \]

2. Move the constant term to the other side:
   \[
   x^2 + \frac{b}{a}x = -\frac{c}{a}
   \]

3. Complete the square:
   - Half of \(\frac{b}{a}\) is \(\frac{b}{2a}\).
   - Square it to get \(\frac{b^2}{4a^2}\).
   - Add \(\frac{b^2}{4a^2}\) to both sides:
     \[
     x^2 + \frac{b}{a}x + \frac{b^2}{4a^2} = -\frac{c}{a} + \frac{b^2}{4a^2}
     \]

4. Simplify the right side:
   \[
   -\frac{c}{a} + \frac{b^2}{4a^2} = \frac{b^2 - 4ac}{4a^2}
   \]

5. Write the left side as a square:
   \[
   \left(x + \frac{b}{2a}\right)^2 = \frac{b^2 - 4ac}{4a^2}
   \]

6. Solve for \( x \) by taking the square root of both sides:
   \[
   x + \frac{b}{2a} = \pm \frac{\sqrt{b^2 - 4ac}}{2a}
   \]

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

This is the quadratic formula.

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Would you like a detailed explanation for any specific part? Here are some related questions for further exploration:

1. What is the purpose of completing the square in solving quadratic equations?
2. How does completing the square relate to the vertex form of a quadratic function?
3. Can all quadratic equations be solved by completing the square?
4. What are the applications of completing the square in real-world problems?
5. How does completing the square help in graphing quadratic functions?

**Tip:** Completing the square can also help to easily identify the maximum or minimum points of a quadratic function, which are essential for understanding its graph.

1. What term can be added to the following equations to make a complete square? a) 0 = x^2 - 8x + ? b) y = x^2 + 9x + ? c) y = x^2 - (b/a)x + ? 2. Solve for x by using the method of completing the square. a) -3x^2 + 5x + 4 = 0 b) ax^2 + bx + c = 0

Learn how to complete the square to solve quadratic equations. This problem involves finding terms to create perfect squares and solving equations using the completing the square method, suitable for students in Grades 9-10.

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