Christian Chester | Portfolium - Free Printable

It seems you've uploaded an image containing a list of algebraic problems, and you're asking for solutions and explanations. Below, I will solve each problem step by step.

---

Problem 1:

$$
\frac{x^2}{3} - \frac{5x}{6} = 0
$$

#### Solution:
1. Find a common denominator for the fractions:
$$
\frac{x^2}{3} - \frac{5x}{6} = 0 \implies \frac{2x^2}{6} - \frac{5x}{6} = 0
$$

2. Combine the terms into a single fraction:
$$
\frac{2x^2 - 5x}{6} = 0
$$

3. Since the denominator cannot be zero, set the numerator equal to zero:
$$
2x^2 - 5x = 0
$$

4. Factor out the greatest common factor (GCF):
$$
x(2x - 5) = 0
$$

5. Use the zero-product property:
$$
x = 0 \quad \text{or} \quad 2x - 5 = 0
$$

6. Solve for \( x \) in the second equation:
$$
2x - 5 = 0 \implies 2x = 5 \implies x = \frac{5}{2}
$$

#### Final Answer:
$$
\boxed{x = 0 \text{ or } x = \frac{5}{2}}
$$

---

Problem 2:

$$
a^2 - 18a + 25 = 0
$$

#### Solution:
This is a quadratic equation. We can solve it using the quadratic formula:
$$
x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}
$$
where \( a = 1 \), \( b = -18 \), and \( c = 25 \).

1. Calculate the discriminant:
$$
b^2 - 4ac = (-18)^2 - 4(1)(25) = 324 - 100 = 224
$$

2. Substitute into the quadratic formula:
$$
a = \frac{-(-18) \pm \sqrt{224}}{2(1)} = \frac{18 \pm \sqrt{224}}{2}
$$

3. Simplify \( \sqrt{224} \):
$$
\sqrt{224} = \sqrt{16 \cdot 14} = 4\sqrt{14}
$$

4. Substitute back:
$$
a = \frac{18 \pm 4\sqrt{14}}{2} = 9 \pm 2\sqrt{14}
$$

#### Final Answer:
$$
\boxed{a = 9 + 2\sqrt{14} \text{ or } a = 9 - 2\sqrt{14}}
$$

---

Problem 3:

$$
(x + 2)(x - 3) = 0
$$

#### Solution:
This is already factored. Use the zero-product property:
$$
x + 2 = 0 \quad \text{or} \quad x - 3 = 0
$$

1. Solve each equation:
$$
x + 2 = 0 \implies x = -2
$$
$$
x - 3 = 0 \implies x = 3
$$

#### Final Answer:
$$
\boxed{x = -2 \text{ or } x = 3}
$$

---

Problem 4:

$$
x^2 + 2x^2 + 3x - 2x - 4 - 6 = 0
$$

#### Solution:
1. Combine like terms:
$$
x^2 + 2x^2 + 3x - 2x - 4 - 6 = 0 \implies 3x^2 + x - 10 = 0
$$

2. This is a quadratic equation. Use the quadratic formula:
$$
x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}
$$
where \( a = 3 \), \( b = 1 \), and \( c = -10 \).

3. Calculate the discriminant:
$$
b^2 - 4ac = (1)^2 - 4(3)(-10) = 1 + 120 = 121
$$

4. Substitute into the quadratic formula:
$$
x = \frac{-1 \pm \sqrt{121}}{2(3)} = \frac{-1 \pm 11}{6}
$$

5. Solve for the two roots:
$$
x = \frac{-1 + 11}{6} = \frac{10}{6} = \frac{5}{3}
$$
$$
x = \frac{-1 - 11}{6} = \frac{-12}{6} = -2
$$

#### Final Answer:
$$
\boxed{x = \frac{5}{3} \text{ or } x = -2}
$$

---

Problem 5:

$$
(3x + 2)(x + 2) = 2x^2
$$

#### Solution:
1. Expand the left-hand side:
$$
(3x + 2)(x + 2) = 3x^2 + 6x + 2x + 4 = 3x^2 + 8x + 4
$$

2. Set the equation equal to \( 2x^2 \):
$$
3x^2 + 8x + 4 = 2x^2
$$

3. Move all terms to one side:
$$
3x^2 + 8x + 4 - 2x^2 = 0 \implies x^2 + 8x + 4 = 0
$$

4. Use the quadratic formula:
$$
x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}
$$
where \( a = 1 \), \( b = 8 \), and \( c = 4 \).

5. Calculate the discriminant:
$$
b^2 - 4ac = (8)^2 - 4(1)(4) = 64 - 16 = 48
$$

6. Substitute into the quadratic formula:
$$
x = \frac{-8 \pm \sqrt{48}}{2(1)} = \frac{-8 \pm \sqrt{16 \cdot 3}}{2} = \frac{-8 \pm 4\sqrt{3}}{2}
$$

7. Simplify:
$$
x = \frac{-8}{2} \pm \frac{4\sqrt{3}}{2} = -4 \pm 2\sqrt{3}
$$

#### Final Answer:
$$
\boxed{x = -4 + 2\sqrt{3} \text{ or } x = -4 - 2\sqrt{3}}
$$

---

Problem 6:

$$
(9x - 3)(5x - 5) = 0
$$

#### Solution:
This is already factored. Use the zero-product property:
$$
9x - 3 = 0 \quad \text{or} \quad 5x - 5 = 0
$$

1. Solve each equation:
$$
9x - 3 = 0 \implies 9x = 3 \implies x = \frac{3}{9} = \frac{1}{3}
$$
$$
5x - 5 = 0 \implies 5x = 5 \implies x = \frac{5}{5} = 1
$$

#### Final Answer:
$$
\boxed{x = \frac{1}{3} \text{ or } x = 1}
$$

---

Problem 7:

$$
x^2 + 5xy + 3y^2 = 0
$$

#### Solution:
This is a quadratic equation in \( x \) with \( y \) as a parameter. Use the quadratic formula:
$$
x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}
$$
where \( a = 1 \), \( b = 5y \), and \( c = 3y^2 \).

1. Calculate the discriminant:
$$
b^2 - 4ac = (5y)^2 - 4(1)(3y^2) = 25y^2 - 12y^2 = 13y^2
$$

2. Substitute into the quadratic formula:
$$
x = \frac{-5y \pm \sqrt{13y^2}}{2(1)} = \frac{-5y \pm y\sqrt{13}}{2}
$$

3. Simplify:
$$
x = \frac{-5y}{2} \pm \frac{y\sqrt{13}}{2} = \frac{-5 \pm \sqrt{13}}{2}y
$$

#### Final Answer:
$$
\boxed{x = \frac{-5 + \sqrt{13}}{2}y \text{ or } x = \frac{-5 - \sqrt{13}}{2}y}
$$

---

Problem 8:

$$
6x^2 + 28xy + y = 0
$$

#### Solution:
This is a quadratic equation in \( x \) with \( y \) as a parameter. Use the quadratic formula:
$$
x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}
$$
where \( a = 6 \), \( b = 28y \), and \( c = y \).

1. Calculate the discriminant:
$$
b^2 - 4ac = (28y)^2 - 4(6)(y) = 784y^2 - 24y
$$

2. Substitute into the quadratic formula:
$$
x = \frac{-28y \pm \sqrt{784y^2 - 24y}}{2(6)} = \frac{-28y \pm \sqrt{784y^2 - 24y}}{12}
$$

#### Final Answer:
$$
\boxed{x = \frac{-28y \pm \sqrt{784y^2 - 24y}}{12}}
$$

---

Problem 9:

$$
x^2 - 3xy + 2y^2 = 0
$$

#### Solution:
This is a quadratic equation in \( x \) with \( y \) as a parameter. Use the quadratic formula:
$$
x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}
$$
where \( a = 1 \), \( b = -3y \), and \( c = 2y^2 \).

1. Calculate the discriminant:
$$
b^2 - 4ac = (-3y)^2 - 4(1)(2y^2) = 9y^2 - 8y^2 = y^2
$$

2. Substitute into the quadratic formula:
$$
x = \frac{-(-3y) \pm \sqrt{y^2}}{2(1)} = \frac{3y \pm y}{2}
$$

3. Solve for the two roots:
$$
x = \frac{3y + y}{2} = \frac{4y}{2} = 2y
$$
$$
x = \frac{3y - y}{2} = \frac{2y}{2} = y
$$

#### Final Answer:
$$
\boxed{x = 2y \text{ or } x = y}
$$

---

Problem 10:

$$
9x^2 - 18xy = 0
$$

#### Solution:
Factor out the greatest common factor (GCF):
$$
9x(x - 2y) = 0
$$

1. Use the zero-product property:
$$
9x = 0 \quad \text{or} \quad x - 2y = 0
$$

2. Solve each equation:
$$
9x = 0 \implies x = 0
$$
$$
x - 2y = 0 \implies x = 2y
$$

#### Final Answer:
$$
\boxed{x = 0 \text{ or } x = 2y}
$$

---

Problem 11:

$$
5x^2 + 3xy - 18y^2 = 0
$$

#### Solution:
This is a quadratic equation in \( x \) with \( y \) as a parameter. Use the quadratic formula:
$$
x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}
$$
where \( a = 5 \), \( b = 3y \), and \( c = -18y^2 \).

1. Calculate the discriminant:
$$
b^2 - 4ac = (3y)^2 - 4(5)(-18y^2) = 9y^2 + 360y^2 = 369y^2
$$

2. Substitute into the quadratic formula:
$$
x = \frac{-3y \pm \sqrt{369y^2}}{2(5)} = \frac{-3y \pm y\sqrt{369}}{10}
$$

3. Simplify:
$$
x = \frac{-3y}{10} \pm \frac{y\sqrt{369}}{10} = \frac{-3 \pm \sqrt{369}}{10}y
$$

#### Final Answer:
$$
\boxed{x = \frac{-3 + \sqrt{369}}{10}y \text{ or } x = \frac{-3 - \sqrt{369}}{10}y}
$$

---

Problem 12:

$$
x^2 - 3xy - 28y^2 = 0
$$

#### Solution:
This is a quadratic equation in \( x \) with \( y \) as a parameter. Use the quadratic formula:
$$
x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}
$$
where \( a = 1 \), \( b = -3y \), and \( c = -28y^2 \).

1. Calculate the discriminant:
$$
b^2 - 4ac = (-3y)^2 - 4(1)(-28y^2) = 9y^2 + 112y^2 = 121y^2
$$

2. Substitute into the quadratic formula:
$$
x = \frac{-(-3y) \pm \sqrt{121y^2}}{2(1)} = \frac{3y \pm 11y}{2}
$$

3. Solve for the two roots:
$$
x = \frac{3y + 11y}{2} = \frac{14y}{2} = 7y
$$
$$
x = \frac{3y - 11y}{2} = \frac{-8y}{2} = -4y
$$

#### Final Answer:
$$
\boxed{x = 7y \text{ or } x = -4y}
$$

---

Problem 13:

$$
x^2 - 3xy - 28y^2 = 0
$$

#### Solution:
This is the same as Problem 12. The solution is:
$$
\boxed{x = 7y \text{ or } x = -4y}
$$

---

Problem 14:

$$
9x^2 - 25 = 0
$$

#### Solution:
1. Factor the difference of squares:
$$
9x^2 - 25 = (3x - 5)(3x + 5) = 0
$$

2. Use the zero-product property:
$$
3x - 5 = 0 \quad \text{or} \quad 3x + 5 = 0
$$

3. Solve each equation:
$$
3x - 5 = 0 \implies 3x = 5 \implies x = \frac{5}{3}
$$
$$
3x + 5 = 0 \implies 3x = -5 \implies x = -\frac{5}{3}
$$

#### Final Answer:
$$
\boxed{x = \frac{5}{3} \text{ or } x = -\frac{5}{3}}
$$

---

Final Boxed Answers:

1. \( \boxed{x = 0 \text{ or } x = \frac{5}{2}} \)
2. \( \boxed{a = 9 + 2\sqrt{14} \text{ or } a = 9 - 2\sqrt{14}} \)
3. \( \boxed{x = -2 \text{ or } x = 3} \)
4. \( \boxed{x = \frac{5}{3} \text{ or } x = -2} \)
5. \( \boxed{x = -4 + 2\sqrt{3} \text{ or } x = -4 - 2\sqrt{3}} \)
6. \( \boxed{x = \frac{1}{3} \text{ or } x = 1} \)
7. \( \boxed{x = \frac{-5 + \sqrt{13}}{2}y \text{ or } x = \frac{-5 - \sqrt{13}}{2}y} \)
8. \( \boxed{x = \frac{-28y \pm \sqrt{784y^2 - 24y}}{12}} \)
9. \( \boxed{x = 2y \text{ or } x = y} \)
10. \( \boxed{x = 0 \text{ or } x = 2y} \)
11. \( \boxed{x = \frac{-3 + \sqrt{369}}{10}y \text{ or } x = \frac{-3 - \sqrt{369}}{10}y} \)
12. \( \boxed{x = 7y \text{ or } x = -4y} \)
13. \( \boxed{x = 7y \text{ or } x = -4y} \)
14. \( \boxed{x = \frac{5}{3} \text{ or } x = -\frac{5}{3}} \)