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### Question 1:
The quadratic equation is given as:
\[ x^2 + 2x + 15 = 0 \]
We are asked to find the correct statement regarding its roots. We can solve it using the discriminant:
\[ \Delta = b^2 - 4ac \]
Where \( a = 1 \), \( b = 2 \), and \( c = 15 \).
\[ \Delta = 2^2 - 4(1)(15) = 4 - 60 = -56 \]

Since the discriminant \( \Delta \) is negative, the quadratic equation has **imaginary roots**, meaning it does not have real solutions.

Answer: **b) Akar-akar PK tidak real (imajiner)**

### Question 2:
One of the roots of the equation \( y^2 + 9y + 2p = 0 \) is given as \( y = -5 \). We need to find the value of \( p \).

Substitute \( y = -5 \) into the equation:
\[
(-5)^2 + 9(-5) + 2p = 0
\]
\[
25 - 45 + 2p = 0
\]
\[
-20 + 2p = 0
\]
\[
2p = 20
\]
\[
p = 10
\]

Answer: \( p = 10 \)

### Question 3:
The quadratic equation is \( 2x^2 + 9x + 4 = 0 \). Let's find the roots using the quadratic formula:
\[
x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}
\]
Where \( a = 2 \), \( b = 9 \), and \( c = 4 \):
\[
x = \frac{-9 \pm \sqrt{9^2 - 4(2)(4)}}{2(2)}
\]
\[
x = \frac{-9 \pm \sqrt{81 - 32}}{4}
\]
\[
x = \frac{-9 \pm \sqrt{49}}{4}
\]
\[
x = \frac{-9 \pm 7}{4}
\]

Thus, the two roots are:
\[
x_1 = \frac{-9 + 7}{4} = \frac{-2}{4} = -\frac{1}{2}
\]
\[
x_2 = \frac{-9 - 7}{4} = \frac{-16}{4} = -4
\]

Answer: **b) \( x_1 = -4 \) atau \( x_2 = -1/2 \)**

### Question 4:
The quadratic equation is \( 6x^2 - x - 2 = 0 \). Let's solve it using the quadratic formula:
\[
x = \frac{-(-1) \pm \sqrt{(-1)^2 - 4(6)(-2)}}{2(6)}
\]
\[
x = \frac{1 \pm \sqrt{1 + 48}}{12}
\]
\[
x = \frac{1 \pm \sqrt{49}}{12}
\]
\[
x = \frac{1 \pm 7}{12}
\]

Thus, the two roots are:
\[
x_1 = \frac{1 + 7}{12} = \frac{8}{12} = \frac{2}{3}
\]
\[
x_2 = \frac{1 - 7}{12} = \frac{-6}{12} = -\frac{1}{2}
\]

Answer: **a) \( \frac{2}{3} \) dan \( -\frac{1}{2} \)**

### Question 5:
We are given that \( \alpha \) and \( \beta \) are the roots of the quadratic equation \( x^2 - 5x - 9 = 0 \). The sum and product of the roots for a quadratic equation \( ax^2 + bx + c = 0 \) are given by:
\[
\alpha + \beta = -\frac{b}{a} \quad \text{and} \quad \alpha \beta = \frac{c}{a}
\]

Here, \( a = 1 \), \( b = -5 \), and \( c = -9 \), so:
\[
\alpha + \beta = -\left( \frac{-5}{1} \right) = 5
\]
\[
\alpha \beta = \frac{-9}{1} = -9
\]

Answer: **a) \( \alpha \beta = -9 \)**

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Would you like more details or explanations on any of these problems?

### Related Questions:
1. How do you solve quadratic equations with imaginary roots?
2. What is the quadratic formula and how is it used in solving quadratic equations?
3. How does the discriminant determine the nature of the roots of a quadratic equation?
4. What are the properties of roots in a quadratic equation?
5. How do you solve word problems involving quadratic equations?

**Tip:** Always check the discriminant when solving quadratic equations to determine the nature of the roots (real, repeated, or imaginary).

The original question is a worksheet with quadratic equations involving various forms like real, imaginary roots, sum/product of roots, and solving for variables.

This page provides a detailed solution to quadratic equations, including determining real and imaginary roots, solving for variables, and using the quadratic formula. Ideal for students in Grades 10-11.

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