Worksheet illustrating the three types of quadratic roots with corresponding graphs and equations.

Problem Overview:

The worksheet introduces the concept of graphing quadratic functions and solving for their roots (zeros). It categorizes quadratic equations based on the number of real roots they have: two real roots, one real root, and no real roots. The task is to solve for the zeros of each quadratic equation provided.

Key Concepts:

1. Parabola: The U-shaped graph of a quadratic function.
2. Roots/Zeros: The x-values where the parabola crosses the x-axis.
3. Types of Roots:
- Two Real Roots: The parabola crosses the x-axis at two distinct points.
- One Real Root: The parabola touches the x-axis at exactly one point (vertex).
- No Real Roots: The parabola does not cross or touch the x-axis.

Given Equations:

1. Two Real Roots: \( x^2 + 2x - 3 = 0 \)
2. One Real Root: \( x^2 + 6x + 9 = 0 \)
3. No Real Roots: \( (x - 1)^2 + 3 = 0 \)

Solution:

#### 1. Solve for the Zeros: \( x^2 + 2x - 3 = 0 \)
This is a quadratic equation in standard form \( ax^2 + bx + c = 0 \), where \( a = 1 \), \( b = 2 \), and \( c = -3 \).

Step 1: Factor the quadratic equation.
We need two numbers that multiply to \( ac = -3 \) and add up to \( b = 2 \). These numbers are \( 3 \) and \( -1 \).

\[
x^2 + 2x - 3 = (x + 3)(x - 1) = 0
\]

Step 2: Solve for \( x \).
Set each factor equal to zero:

\[
x + 3 = 0 \quad \text{or} \quad x - 1 = 0
\]

\[
x = -3 \quad \text{or} \quad x = 1
\]

Solution: The zeros are \( x = -3 \) and \( x = 1 \).

#### 2. Solve for the Zeros: \( x^2 + 6x + 9 = 0 \)
This is a quadratic equation in standard form \( ax^2 + bx + c = 0 \), where \( a = 1 \), \( b = 6 \), and \( c = 9 \).

Step 1: Factor the quadratic equation.
Notice that this is a perfect square trinomial:

\[
x^2 + 6x + 9 = (x + 3)^2 = 0
\]

Step 2: Solve for \( x \).
Set the factor equal to zero:

\[
(x + 3)^2 = 0
\]

\[
x + 3 = 0
\]

\[
x = -3
\]

Solution: The zero is \( x = -3 \) (a repeated root).

#### 3. Solve for the Zeros: \( (x - 1)^2 + 3 = 0 \)
This equation is already in a simplified form.

Step 1: Isolate the squared term.
Subtract 3 from both sides:

\[
(x - 1)^2 = -3
\]

Step 2: Analyze the equation.
The left side, \( (x - 1)^2 \), is a square term and is always non-negative (i.e., \( (x - 1)^2 \geq 0 \)). However, the right side is \(-3\), which is negative. This means there are no real values of \( x \) that satisfy the equation.

Solution: There are no real roots.

Final Answers:

1. Two Real Roots: \( x^2 + 2x - 3 = 0 \)
\[
\boxed{x = -3, x = 1}
\]

2. One Real Root: \( x^2 + 6x + 9 = 0 \)
\[
\boxed{x = -3}
\]

3. No Real Roots: \( (x - 1)^2 + 3 = 0 \)
\[
\boxed{\text{No real roots}}
\]