College Algebra Study Guide: Quadratic, Radical, and Rational Equations

Q1. Geometry Application: Diagonal of a Rectangular Park

Background

Topic: Quadratic Equations & Geometry

This question tests your ability to use the Pythagorean Theorem to find the length of the diagonal in a rectangle.

Key Terms and Formulas

• Pythagorean Theorem: $a^2 + b^2 = c^2$

• Where $a$ and $b$ are the sides of the rectangle, and $c$ is the diagonal.

Step-by-Step Guidance

1. Identify the sides of the rectangle: $a = 4$ miles, $b = 2$ miles.

2. Set up the Pythagorean Theorem: $c^2 = a^2 + b^2$.

3. Substitute the values: $c^2 = 4^2 + 2^2$.

4. Calculate $4^2$ and $2^2$ separately, then add them together.

5. Take the square root to solve for $c$ (the diagonal).

Try solving on your own before revealing the answer!

Final Answer: $2\sqrt{5}$ miles (exact), approximately 4.5 miles (rounded to the nearest tenth)

Using the Pythagorean Theorem: $c = \sqrt{4^2 + 2^2} = \sqrt{16 + 4} = \sqrt{20} = 2\sqrt{5} \approx 4.5$ miles.

The diagonal is $2\sqrt{5}$ miles, which is about 4.5 miles when rounded.

Q2. Geometry Application: Dimensions of a Rectangular Patio

Background

Topic: Quadratic Equations & Area

This question tests your ability to set up and solve a quadratic equation based on area and relationships between length and width.

Key Terms and Formulas

• Area of rectangle: $A = l \times w$

• Width is 8 feet longer than length: $w = l + 8$

Step-by-Step Guidance

1. Let $l$ be the length. Then $w = l + 8$.

2. Set up the area equation: $l \times (l + 8) = 180$.

3. Expand and rearrange: $l^2 + 8l = 180$.

4. Move all terms to one side: $l^2 + 8l - 180 = 0$.

5. Set up to solve the quadratic equation for $l$ (length).

Try solving on your own before revealing the answer!

Final Answer: Length = 10 ft, Width = 18 ft

Solving $l^2 + 8l - 180 = 0$ gives $l = 10$ (since length must be positive), so width is $10 + 8 = 18$ ft.

The patio's dimensions are 10 feet by 18 feet.

Q3. Tuition Model: Approximate Cost in 2000

Background

Topic: Quadratic Models

This question tests your ability to evaluate a quadratic function for a given value of $x$.

Key Terms and Formulas

• Quadratic model: $y = 18.34x^2 + 512.9x + 9314$

• $x = 0$ corresponds to the year 2000.

Step-by-Step Guidance

1. Identify the value of $x$ for 2000: $x = 0$.

2. Substitute $x = 0$ into the model: $y = 18.34(0)^2 + 512.9(0) + 9314$.

3. Simplify each term and add them together.

Try solving on your own before revealing the answer!

Final Answer: $y = 9314$

For $x = 0$, the cost is $9314$ dollars.

This is the base value in the model for the year 2000.

Q4. Tuition Model: Approximate Cost in 2009

Background

Topic: Quadratic Models

This question tests your ability to evaluate a quadratic function for a specific year.

Key Terms and Formulas

• Quadratic model: $y = 18.34x^2 + 512.9x + 9314$

• $x = 9$ corresponds to the year 2009.

Step-by-Step Guidance

1. Identify the value of $x$ for 2009: $x = 9$.

2. Substitute $x = 9$ into the model: $y = 18.34(9)^2 + 512.9(9) + 9314$.

3. Calculate $9^2$ and multiply by $18.34$.

4. Multiply $512.9$ by $9$.

5. Add all terms together to find $y$.

Try solving on your own before revealing the answer!

Final Answer: $y \approx 11702$

Plugging in $x = 9$ gives $y = 18.34 \times 81 + 512.9 \times 9 + 9314 \approx 11702$ dollars.

This is the approximate cost in 2009.

Q5. Tuition Model: Year When Cost Was $11702

Background

Topic: Quadratic Models & Solving Equations

This question tests your ability to solve a quadratic equation for $x$ given a value of $y$.

Key Terms and Formulas

• Quadratic model: $y = 18.34x^2 + 512.9x + 9314$

• Set $y = 11702$ and solve for $x$.

Step-by-Step Guidance

1. Set up the equation: $11702 = 18.34x^2 + 512.9x + 9314$.

2. Subtract $9314$ from both sides to isolate the quadratic terms.

3. Rearrange to standard quadratic form: $18.34x^2 + 512.9x + (9314 - 11702) = 0$.

4. Simplify the constant term and write the equation as $18.34x^2 + 512.9x - 2388 = 0$.

5. Set up to solve for $x$ using the quadratic formula.

Try solving on your own before revealing the answer!

Final Answer: $x \approx 9$ (Year 2009)

Solving $18.34x^2 + 512.9x - 2388 = 0$ gives $x \approx 9$, which corresponds to the year 2009.

This matches the previous calculation for the cost in 2009.

Q6. Radical Equation: $x - 3 = 2$

Background

Topic: Radical Equations

This question tests your ability to solve a simple radical equation and check for extraneous solutions.

Key Terms and Formulas

• Radical equation: An equation involving a square root or other root.

• Extraneous solution: A solution that does not satisfy the original equation.

Step-by-Step Guidance

1. Isolate the radical: $\sqrt{x - 3} = 2$.

2. Square both sides to eliminate the radical: $(\sqrt{x - 3})^2 = 2^2$.

3. Simplify: $x - 3 = 4$.

4. Solve for $x$ by adding $3$ to both sides.

5. Check the solution in the original equation to ensure it is not extraneous.

Try solving on your own before revealing the answer!

Final Answer: Solution set: $\{7\}$

Solving gives $x = 7$. Checking: $\sqrt{7 - 3} = \sqrt{4} = 2$, which matches the original equation.

So $x = 7$ is a valid solution.

Q7. Radical Equation: $2x + 1 = 5$

Background

Topic: Radical Equations

This question tests your ability to solve a radical equation and check for extraneous solutions.

Key Terms and Formulas

• Radical equation: $\sqrt{2x + 1} = 5$

• Extraneous solution: Always check your answer in the original equation.

Step-by-Step Guidance

1. Isolate the radical: $\sqrt{2x + 1} = 5$.

2. Square both sides: $(\sqrt{2x + 1})^2 = 5^2$.

3. Simplify: $2x + 1 = 25$.

4. Solve for $x$ by subtracting $1$and dividing by $2$.

5. Check the solution in the original equation for extraneous solutions.

Try solving on your own before revealing the answer!

Final Answer: Solution set: $\{12\}$

Solving gives $x = 12$. Checking: $\sqrt{2 \times 12 + 1} = \sqrt{25} = 5$, which matches the original equation.

So $x = 12$ is a valid solution.

Q8. Radical Equation: $\sqrt{x^2 - 2x} = x$

Background

Topic: Radical Equations

This question tests your ability to solve a radical equation where the variable appears both inside and outside the radical.

Key Terms and Formulas

• Radical equation: $\sqrt{x^2 - 2x} = x$

• Extraneous solution: Always check your answer in the original equation.

Step-by-Step Guidance

1. Square both sides: $(\sqrt{x^2 - 2x})^2 = x^2$.

2. Simplify: $x^2 - 2x = x^2$.

3. Subtract $x^2$ from both sides: $x^2 - 2x - x^2 = 0$.

4. Simplify further and solve for $x$.

5. Check for extraneous solutions by substituting back into the original equation.

Try solving on your own before revealing the answer!

Final Answer: Solution set: $\{0\}$

Solving gives $x = 0$. Checking: $\sqrt{0^2 - 2 \times 0} = \sqrt{0} = 0$, which matches the original equation.

So $x = 0$ is a valid solution.

Q9. Radical Equation: $\sqrt{x + 7} = x + 1$

Background

Topic: Radical Equations

This question tests your ability to solve a radical equation and check for extraneous solutions.

Key Terms and Formulas

• Radical equation: $\sqrt{x + 7} = x + 1$

• Extraneous solution: Always check your answer in the original equation.

Step-by-Step Guidance

1. Square both sides: $(\sqrt{x + 7})^2 = (x + 1)^2$.

2. Simplify: $x + 7 = x^2 + 2x + 1$.

3. Move all terms to one side: $x^2 + 2x + 1 - x - 7 = 0$.

4. Simplify and set up the quadratic equation to solve for $x$.

5. Check for extraneous solutions by substituting back into the original equation.

Try solving on your own before revealing the answer!

Final Answer: Solution set: $\{2\}$

Solving gives $x = 2$. Checking: $\sqrt{2 + 7} = \sqrt{9} = 3$, and $2 + 1 = 3$, which matches the original equation.

So $x = 2$ is a valid solution.

Q10. Rational Equation: $\frac{1}{x} = \frac{1}{5} + \frac{3}{2x}$

Background

Topic: Rational Equations

This question tests your ability to solve rational equations, determine the domain, and find the least common denominator (LCD).

Key Terms and Formulas

• Rational equation: An equation involving fractions with variables in the denominator.

• Domain: Values of $x$ for which the denominators are not zero.

• LCD: Least common denominator for all fractions.

Step-by-Step Guidance

1. Identify the domain: $x \neq 0$.

2. Find the LCD for $x$, $5$, and $2x$; the LCD is $10x$.

3. Multiply both sides of the equation by the LCD to clear denominators.

4. Simplify the resulting equation and solve for $x$.

5. Check for extraneous solutions by ensuring $x$ is not zero.

Try solving on your own before revealing the answer!

Final Answer: Solution set: $\{2\}$

Solving gives $x = 2$. The domain excludes $x = 0$, so $x = 2$ is valid.

Q11. Rational Equation: $\frac{8}{x+3} = \frac{5}{x-2}$

Background

Topic: Rational Equations

This question tests your ability to solve rational equations and determine the domain and LCD.

Key Terms and Formulas

• Domain: $x \neq -3$, $x \neq 2$ (denominators cannot be zero).

• LCD: $(x+3)(x-2)$

Step-by-Step Guidance

1. Identify the domain: $x \neq -3$, $x \neq 2$.

2. Multiply both sides by the LCD: $(x+3)(x-2)$.

3. Simplify the resulting equation and solve for $x$.

4. Check for extraneous solutions by ensuring $x$ is not $-3$ or $2$.

Try solving on your own before revealing the answer!

Final Answer: Solution set: $\{7\}$

Solving gives $x = 7$. The domain excludes $x = -3$ and $x = 2$, so $x = 7$ is valid.

Q12. Rational Equation: $\frac{3}{x} + \frac{x-2}{8x} = \frac{13}{16}$

Background

Topic: Rational Equations

This question tests your ability to solve rational equations, determine the domain, and find the LCD.

Key Terms and Formulas

• Domain: $x \neq 0$

• LCD: $16x$

Step-by-Step Guidance

1. Identify the domain: $x \neq 0$.

2. Multiply both sides by the LCD: $16x$.

3. Simplify the resulting equation and solve for $x$.

4. Check for extraneous solutions by ensuring $x$ is not zero.

Try solving on your own before revealing the answer!

Final Answer: Solution set: $\{4\}$

Solving gives $x = 4$. The domain excludes $x = 0$, so $x = 4$ is valid.

Q13. Rational Equation: $\frac{4}{x+3} - 2 = \frac{5x-9}{x+3}$

Background

Topic: Rational Equations

This question tests your ability to solve rational equations, determine the domain, and find the LCD.

Key Terms and Formulas

• Domain: $x \neq -3$

• LCD: $x+3$

Step-by-Step Guidance

1. Identify the domain: $x \neq -3$.

2. Multiply both sides by the LCD: $x+3$.

3. Simplify the resulting equation and solve for $x$.

4. Check for extraneous solutions by ensuring $x$ is not $-3$.

Try solving on your own before revealing the answer!

Final Answer: Solution set: $\{1\}$

Solving gives $x = 1$. The domain excludes $x = -3$, so $x = 1$ is valid.