Math worksheet with matching quadratic equations and word problems requiring solutions using the method of extracting square roots.

Part II: Matching Type

#### Directions:
Column B contains the roots of the quadratic equations written in Column A. Solve each quadratic equation using the method of extracting square roots. Write the letter of your choice on the space provided before the number in Column A. You may choose the answer not more than once.

#### Column A: Quadratic Equations
1. \( 6x^2 - 49 = 0 \)
2. \( 4x^2 + 2 = 0 \)
3. \( 25x^2 - 36 = 0 \)
4. \( (x + 5)^2 = 16 \)
5. \( 10(3x + 4)^2 = 9 \)
6. \( (x - 6)^2 = 5 \)
7. \( 2(x + 3)^2 = 8 \)

#### Column B: Roots
A. \( x \notin \mathbb{R} \)
B. \( x_1 = -\frac{7}{3}, x_2 = -\frac{1}{3} \)
C. \( x_1 = -9, x_2 = -1 \)
D. \( x_1 = 6 - \sqrt{5}, x_2 = 6 + \sqrt{5} \)
E. \( x_1 = -\frac{7}{4}, x_2 = \frac{7}{4} \)
F. \( x_1 = -5, x_2 = -1 \)
G. \( x_1 = -\frac{6}{5}, x_2 = \frac{6}{5} \)
H. \( x = 0 \)

---

Solutions for Column A:

#### 1. \( 6x^2 - 49 = 0 \)
\[
6x^2 = 49
\]
\[
x^2 = \frac{49}{6}
\]
\[
x = \pm \sqrt{\frac{49}{6}} = \pm \frac{7}{\sqrt{6}} = \pm \frac{7\sqrt{6}}{6}
\]
This does not match any option directly, so we need to recheck or approximate. However, none of the given options match this result exactly. Let's move to the next equation.

#### 2. \( 4x^2 + 2 = 0 \)
\[
4x^2 = -2
\]
\[
x^2 = -\frac{2}{4} = -\frac{1}{2}
\]
Since \( x^2 \) cannot be negative for real numbers, the roots are not real.
\[
\boxed{A}
\]

#### 3. \( 25x^2 - 36 = 0 \)
\[
25x^2 = 36
\]
\[
x^2 = \frac{36}{25}
\]
\[
x = \pm \sqrt{\frac{36}{25}} = \pm \frac{6}{5}
\]
\[
\boxed{G}
\]

#### 4. \( (x + 5)^2 = 16 \)
\[
x + 5 = \pm \sqrt{16}
\]
\[
x + 5 = \pm 4
\]
\[
x = -5 \pm 4
\]
\[
x_1 = -5 + 4 = -1, \quad x_2 = -5 - 4 = -9
\]
\[
\boxed{F}
\]

#### 5. \( 10(3x + 4)^2 = 9 \)
\[
(3x + 4)^2 = \frac{9}{10}
\]
\[
3x + 4 = \pm \sqrt{\frac{9}{10}}
\]
\[
3x + 4 = \pm \frac{3}{\sqrt{10}} = \pm \frac{3\sqrt{10}}{10}
\]
\[
3x = -4 \pm \frac{3\sqrt{10}}{10}
\]
\[
x = \frac{-4 \pm \frac{3\sqrt{10}}{10}}{3} = \frac{-40 \pm 3\sqrt{10}}{30} = \frac{-40}{30} \pm \frac{3\sqrt{10}}{30} = -\frac{4}{3} \pm \frac{\sqrt{10}}{10}
\]
This does not match any option directly. Let's move to the next equation.

#### 6. \( (x - 6)^2 = 5 \)
\[
x - 6 = \pm \sqrt{5}
\]
\[
x = 6 \pm \sqrt{5}
\]
\[
x_1 = 6 - \sqrt{5}, \quad x_2 = 6 + \sqrt{5}
\]
\[
\boxed{D}
\]

#### 7. \( 2(x + 3)^2 = 8 \)
\[
(x + 3)^2 = \frac{8}{2} = 4
\]
\[
x + 3 = \pm \sqrt{4}
\]
\[
x + 3 = \pm 2
\]
\[
x = -3 \pm 2
\]
\[
x_1 = -3 + 2 = -1, \quad x_2 = -3 - 2 = -5
\]
\[
\boxed{F}
\]

---

Final Answers for Part II:

1. \( \boxed{A} \)
2. \( \boxed{G} \)
3. \( \boxed{F} \)
4. \( \boxed{D} \)
5. Not matched (approximation needed)
6. \( \boxed{F} \)

---

Part III: Word Problems

#### 13. Julia drops her camera in the river from a bridge that is 400 feet high. How long is it before she hears the splash?

The time it takes for the camera to fall can be calculated using the free-fall formula:
\[
h = \frac{1}{2}gt^2
\]
where \( h = 400 \) feet, \( g = 32 \) ft/s² (acceleration due to gravity on Earth).

\[
400 = \frac{1}{2} \cdot 32 \cdot t^2
\]
\[
400 = 16t^2
\]
\[
t^2 = \frac{400}{16} = 25
\]
\[
t = \sqrt{25} = 5 \text{ seconds}
\]

Thus, the time before Julia hears the splash is:
\[
\boxed{5 \text{ seconds}}
\]

#### 14. It takes a rock 5.3 seconds to splash in the water when it is dropped from the top of a cliff. How high is the cliff in meters?

Using the free-fall formula again:
\[
h = \frac{1}{2}gt^2
\]
where \( t = 5.3 \) seconds and \( g = 9.8 \) m/s² (acceleration due to gravity on Earth).

\[
h = \frac{1}{2} \cdot 9.8 \cdot (5.3)^2
\]
\[
h = 4.9 \cdot 28.09
\]
\[
h \approx 137.64 \text{ meters}
\]

Thus, the height of the cliff is:
\[
\boxed{137.64 \text{ meters}}
\]

#### 15. Jayssa drops a rock from the roof of a building 50 feet high. Jonelle drops a quarter from the top story window, 40 feet high, exactly half a second after Jayssa drops the rock. Which hits the ground first?

Jayssa's Rock:
\[
h = \frac{1}{2}gt^2
\]
\[
50 = \frac{1}{2} \cdot 32 \cdot t^2
\]
\[
50 = 16t^2
\]
\[
t^2 = \frac{50}{16} = 3.125
\]
\[
t = \sqrt{3.125} \approx 1.77 \text{ seconds}
\]

Jonelle's Quarter:
\[
h = \frac{1}{2}gt^2
\]
\[
40 = \frac{1}{2} \cdot 32 \cdot t^2
\]
\[
40 = 16t^2
\]
\[
t^2 = \frac{40}{16} = 2.5
\]
\[
t = \sqrt{2.5} \approx 1.58 \text{ seconds}
\]

However, Jonelle drops the quarter 0.5 seconds after Jayssa, so the total time for Jonelle's quarter is:
\[
1.58 + 0.5 = 2.08 \text{ seconds}
\]

Comparing the times:
- Jayssa's rock: \( 1.77 \) seconds
- Jonelle's quarter: \( 2.08 \) seconds

Thus, Jayssa's rock hits the ground first.

\[
\boxed{\text{Jayssa's rock}}
\]

---

Final Answers for Part III:

13. \( \boxed{5 \text{ seconds}} \)
14. \( \boxed{137.64 \text{ meters}} \)
15. \( \boxed{\text{Jayssa's rock}} \)