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View the Desmos activity to analyze the 4 functions that will be used in this lesson.

[tex]\[
\begin{array}{l}
f_1(x) = (x^2 + 4x + 6) - (2x + 9) \\
g(x) = (x^2 - 5x + 3) + (x + 1) \\
f_2(x) = (x^2 + 4x + 2) - (2x - 5) \\
f_3(x) = (x^2 - 3x + 2) - (2x - 3)
\end{array}
\][/tex]

Question 1

Match the correct roots to each function. Some functions will be used more than once if they have multiple roots.

[tex]\[
\begin{array}{ll}
(-3, 0) & \square \quad \square \\
\left(\frac{5 - \sqrt{5}}{2}, 0\right) & \text{[Select]} \quad \square \\
\text{No Real Roots} & \text{[Select]} \quad \square \\
(2, 0) & \square \quad \text{[Select]} \\
(1, 0) & \text{[Select]} \quad \square \\
\left(\frac{5 + \sqrt{5}}{2}, 0\right) & \text{[Select]} \quad \square \\
\end{array}
\][/tex]

Sagot :

GinnyAnswer
To determine the correct roots for each function, we will solve each equation where the function equals zero, step-by-step, and match them accordingly.

### Step 1: Simplifying Functions

1. [tex]\( f_1(x) \)[/tex]:
[tex]\[ f_1(x) = (x^2 + 4x + 6) - (2x + 9) = x^2 + 4x + 6 - 2x - 9 = x^2 + 2x - 3 \][/tex]
Solve [tex]\( x^2 + 2x - 3 = 0 \)[/tex].

2. [tex]\( g(x) \)[/tex]:
[tex]\[ g(x) = (x^2 - 5x + 3) + (x + 1) = x^2 - 5x + 3 + x + 1 = x^2 - 4x + 4 \][/tex]
Solve [tex]\( x^2 - 4x + 4 = 0 \)[/tex].

3. [tex]\( f_2(x) \)[/tex]:
[tex]\[ f_2(x) = (x^2 + 4x + 2) - (2x - 5) = x^2 + 4x + 2 - 2x + 5 = x^2 + 2x + 7 \][/tex]
Solve [tex]\( x^2 + 2x + 7 = 0 \)[/tex].

4. [tex]\( f_3(x) \)[/tex]:
[tex]\[ f_3(x) = (x^2 - 3x + 2) - (2x - 3) = x^2 - 3x + 2 - 2x + 3 = x^2 - 5x + 5 \][/tex]
Solve [tex]\( x^2 - 5x + 5 = 0 \)[/tex].

### Step 2: Solving the Equations

1. For [tex]\( f_1(x) = x^2 + 2x - 3 = 0 \)[/tex]:
[tex]\[ (x + 3)(x - 1) = 0 \][/tex]
Solutions: [tex]\( x = -3 \)[/tex], [tex]\( x = 1 \)[/tex]

2. For [tex]\( g(x) = x^2 - 4x + 4 = 0 \)[/tex]:
[tex]\[ (x - 2)^2 = 0 \][/tex]
Solution: [tex]\( x = 2 \)[/tex]

3. For [tex]\( f_2(x) = x^2 + 2x + 7 = 0 \)[/tex]:
The discriminant [tex]\((2)^2 - 4 \cdot 1 \cdot 7 = 4 - 28 = -24 \)[/tex], which is less than 0.
Solution: No real roots

4. For [tex]\( f_3(x) = x^2 - 5x + 5 = 0 \)[/tex]:
[tex]\[ x = \frac{5 \pm \sqrt{5}}{2} \][/tex]
Solutions: [tex]\( x = \frac{5 - \sqrt{5}}{2} \)[/tex], [tex]\( x = \frac{5 + \sqrt{5}}{2} \)[/tex]

### Step 3: Matching the Correct Roots

1. [tex]\( f_1(x) = x^2 + 2x - 3 \)[/tex]:
- Roots: [tex]\( (-3, 0) \)[/tex], [tex]\( (1, 0) \)[/tex]

2. [tex]\( g(x) = x^2 - 4x + 4 \)[/tex]:
- Root: [tex]\( (2, 0) \)[/tex]

3. [tex]\( f_2(x) = x^2 + 2x + 7 \)[/tex]:
- No real roots

4. [tex]\( f_3(x) = x^2 - 5x + 5 \)[/tex]:
- Roots: [tex]\( \left( \frac{5 - \sqrt{5}}{2}, 0 \right) \)[/tex], [tex]\( \left( \frac{5 + \sqrt{5}}{2}, 0 \right) \)[/tex]

### Final Answer:

- [tex]\((-3, 0)\)[/tex]:
- [tex]\(f_1(x)\)[/tex]

- [tex]\(\left(\frac{5-\sqrt{5}}{2}, 0\right)\)[/tex]:
- [tex]\(f_3(x)\)[/tex]

- No Real Roots:
- [tex]\(f_2(x)\)[/tex]

- [tex]\((2, 0)\)[/tex]:
- [tex]\(g(x)\)[/tex]

- [tex]\((1, 0)\)[/tex]:
- [tex]\(f_1(x)\)[/tex]

- [tex]\(\left(\frac{5+\sqrt{5}}{2}, 0\right)\)[/tex]:
- [tex]\(f_3(x)\)[/tex]

These roots have been matched correctly in accordance with their respective functions.
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