College Algebra Study Guide: Rational and Radical Equations

Q1. What values of the variable cannot possibly be solutions for the given equation, without actually solving the equation?

$\frac{8}{3x+2} - \frac{1}{x-7} = 0$

Background

Topic: Rational Equations & Domain Restrictions

This question tests your understanding of rational equations and the concept of excluded values (values that make the denominator zero).

Key Terms and Formulas:

• Rational Equation: An equation involving fractions with polynomials in the numerator and denominator.

• Domain Restriction: Values that make any denominator zero are not allowed as solutions.

Step-by-Step Guidance

1. Identify the denominators in the equation: $3x+2$ and $x-7$.

2. Set each denominator equal to zero to find the values that are not allowed: $3x+2=0$ and $x-7=0$.

3. Solve each equation for $x$ to find the excluded values.

Try solving on your own before revealing the answer!

Final Answer: The solutions cannot include $x = -\frac{2}{3}, 7$.

These values make the denominators zero, so they must be excluded from the solution set.

Q2. Decide what values of the variable cannot possibly be solutions for the equation. Do not solve.

$\frac{1}{x-4} + \frac{1}{x+1} = \frac{1}{x^2-3x-4}$

Background

Topic: Rational Equations & Domain Restrictions

This question is about identifying values that make any denominator zero, which are not valid solutions.

Key Terms and Formulas:

• Excluded Values: Values that make denominators zero.

• Factoring: Factor $x^2-3x-4$ to find additional restrictions.

Step-by-Step Guidance

1. Identify all denominators: $x-4$, $x+1$, and $x^2-3x-4$.

2. Factor $x^2-3x-4$ to get $(x-4)(x+1)$.

3. Set each denominator equal to zero and solve for $x$.

Try solving on your own before revealing the answer!

Final Answer: The solutions cannot include $x = 4, -1$.

These values make the denominators zero and must be excluded.

Q3. Solve the equation:

$\frac{2x+4}{2} - \frac{3x}{x-2} = x$

Background

Topic: Rational Equations

This question tests your ability to solve rational equations and check for extraneous solutions.

Key Terms and Formulas:

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

• Extraneous Solution: A solution that does not satisfy the original equation due to domain restrictions.

Step-by-Step Guidance

1. Find a common denominator for all terms.

2. Multiply both sides by the common denominator to clear fractions.

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

Try solving on your own before revealing the answer!

Final Answer: The solution set is $\{-4\}$.

After solving and checking for extraneous solutions, $x = -4$ is valid.

Q4. Solve the equation:

$\frac{6}{x^2-1} - \frac{3}{x-1} = \frac{7}{x+1}$

Background

Topic: Rational Equations

This question tests your ability to solve rational equations and check for extraneous solutions.

Key Terms and Formulas:

• Factoring: $x^2-1 = (x-1)(x+1)$

• Extraneous Solution: Solutions that make denominators zero must be excluded.

Step-by-Step Guidance

1. Factor denominators where possible.

2. Find a common denominator and multiply through to clear fractions.

3. Simplify and solve for $x$.

Try solving on your own before revealing the answer!

Final Answer: The solution set is empty.

No values of $x$ satisfy the equation without making a denominator zero.

Q5. Solve the equation:

$\frac{2x+1}{x-2} + \frac{4}{x} = \frac{-8}{x^2-2x}$

Background

Topic: Rational Equations

This question tests your ability to solve rational equations and check for extraneous solutions.

Key Terms and Formulas:

• Factoring: $x^2-2x = x(x-2)$

• Extraneous Solution: Solutions that make denominators zero must be excluded.

Step-by-Step Guidance

1. Factor denominators where possible.

2. Find a common denominator and multiply through to clear fractions.

3. Simplify and solve for $x$.

Try solving on your own before revealing the answer!

Final Answer: The solution set is $\left\{-\frac{5}{2}\right\}$.

After solving and checking for extraneous solutions, $x = -\frac{5}{2}$ is valid.

Q6. Solve the equation:

$4 = \frac{7}{4x-3} - \frac{3}{(4x-3)^2}$

Background

Topic: Rational Equations

This question tests your ability to solve rational equations and check for extraneous solutions.

Key Terms and Formulas:

• Common Denominator: $(4x-3)^2$

• Extraneous Solution: Solutions that make denominators zero must be excluded.

Step-by-Step Guidance

1. Multiply both sides by $(4x-3)^2$ to clear fractions.

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

Try solving on your own before revealing the answer!

Final Answer: The solution set is $\left\{\frac{15}{16}, 1\right\}$.

Both values satisfy the equation and do not make the denominator zero.

Q7. Solve the equation:

$x - \sqrt{35-2x} = 0$

Background

Topic: Radical Equations

This question tests your ability to solve equations involving square roots and check for extraneous solutions.

Key Terms and Formulas:

• Radical Equation: An equation with a variable inside a square root.

• Extraneous Solution: Solutions that do not satisfy the original equation.

Step-by-Step Guidance

1. Isolate the square root term.

2. Square both sides to eliminate the radical.

3. Solve the resulting quadratic equation for $x$.

Try solving on your own before revealing the answer!

Final Answer: The solution set is $\{5\}$.

After checking, $x = 5$ is valid and does not produce a negative under the square root.

Q8. Solve the equation:

$\sqrt{x-6} = 1 = 3x-23$

Background

Topic: Radical Equations

This question tests your ability to solve equations involving square roots and check for extraneous solutions.

Key Terms and Formulas:

• Radical Equation: An equation with a variable inside a square root.

• Extraneous Solution: Solutions that do not satisfy the original equation.

Step-by-Step Guidance

1. Isolate the square root term.

2. Square both sides to eliminate the radical.

3. Solve the resulting equation for $x$.

Try solving on your own before revealing the answer!

Final Answer: The solution set is $\{10\}$.

After checking, $x = 10$ is valid and does not produce a negative under the square root.

Q9. Solve the equation:

$\sqrt{4y+20} - \sqrt{y-4} = 6$

Background

Topic: Radical Equations

This question tests your ability to solve equations involving square roots and check for extraneous solutions.

Key Terms and Formulas:

• Radical Equation: An equation with a variable inside a square root.

• Extraneous Solution: Solutions that do not satisfy the original equation.

Step-by-Step Guidance

1. Isolate one of the square root terms.

2. Square both sides to eliminate the radical.

3. Solve the resulting equation for $y$.

Try solving on your own before revealing the answer!

Final Answer: The solution set is $\{4, 20\}$.

Both values satisfy the equation and do not produce a negative under the square root.

Q10. Solve the equation:

$\sqrt{3x} = \sqrt{7x+16} - 4$

Background

Topic: Radical Equations

This question tests your ability to solve equations involving square roots and check for extraneous solutions.

Key Terms and Formulas:

• Radical Equation: An equation with a variable inside a square root.

• Extraneous Solution: Solutions that do not satisfy the original equation.

Step-by-Step Guidance

1. Isolate one of the square root terms.

2. Square both sides to eliminate the radical.

3. Solve the resulting equation for $x$.

Try solving on your own before revealing the answer!

Final Answer: The solution set is $\{0, 12\}$.

Both values satisfy the equation and do not produce a negative under the square root.

Q11. Solve the equation:

$(x-4)^{2/3} = 4$

Background

Topic: Radical and Rational Exponents

This question tests your ability to solve equations with rational exponents.

Key Terms and Formulas:

• Rational Exponent: $a^{m/n}$ means the $n$th root of $a$ raised to the $m$th power.

• Solving: Raise both sides to the reciprocal exponent to isolate $x$.

Step-by-Step Guidance

1. Raise both sides to the $\frac{3}{2}$ power to isolate $x-4$.

2. Solve for $x$.

Try solving on your own before revealing the answer!

Final Answer: The solution set is $\{-4, 12\}$.

Both values satisfy the equation after checking for extraneous solutions.

Q12. Solve the equation:

$(2x-3)^{2/3} = x^{1/3}$

Background

Topic: Radical and Rational Exponents

This question tests your ability to solve equations with rational exponents.

Key Terms and Formulas:

• Rational Exponent: $a^{m/n}$ means the $n$th root of $a$ raised to the $m$th power.

• Solving: Raise both sides to the appropriate power to isolate $x$.

Step-by-Step Guidance

1. Raise both sides to the $3$rd power to eliminate the rational exponents.

2. Solve for $x$.

Try solving on your own before revealing the answer!

Final Answer: The solution set is $\left\{1, \frac{9}{4}\right\}$.

Both values satisfy the equation after checking for extraneous solutions.

Q13. Solve the equation:

$15x^4 - 14x^2 + 3 = 0$

Background

Topic: Polynomial Equations

This question tests your ability to solve quartic equations by factoring or substitution.

Key Terms and Formulas:

• Quartic Equation: An equation of degree 4.

• Substitution: Let $y = x^2$ to reduce the degree.

Step-by-Step Guidance

1. Let $y = x^2$ and rewrite the equation in terms of $y$.

2. Solve the resulting quadratic equation for $y$.

3. Take square roots to solve for $x$.

Try solving on your own before revealing the answer!

Final Answer: The solution set is $\left\{\frac{\sqrt{15}}{5}, \frac{\sqrt{3}}{3}, -\frac{\sqrt{3}}{3}, -\frac{\sqrt{15}}{5}\right\}$.

These values are found by solving the quadratic and taking square roots.

Q14. Solve the equation:

$(x-1)^{2/3} + (x-1)^{1/3} - 20 = 0$

Background

Topic: Radical and Rational Exponents

This question tests your ability to solve equations with rational exponents using substitution.

Key Terms and Formulas:

• Substitution: Let $y = (x-1)^{1/3}$ to reduce the equation.

• Solving: Solve the resulting quadratic equation for $y$.

Step-by-Step Guidance

1. Let $y = (x-1)^{1/3}$ and rewrite the equation in terms of $y$.

2. Solve the resulting quadratic equation for $y$.

3. Raise both sides to the 3rd power to solve for $x$.

Try solving on your own before revealing the answer!

Final Answer: The solution set is $\{-124, 65\}$.

Both values satisfy the equation after checking for extraneous solutions.

Q15. Solve the equation:

$15(x^2)^4 - 20(x^2)^2 + 2x^2 = -4$

Background

Topic: Polynomial Equations

This question tests your ability to solve quartic equations by factoring or substitution.

Key Terms and Formulas:

• Quartic Equation: An equation of degree 4.

• Substitution: Let $y = x^2$ to reduce the degree.

Step-by-Step Guidance

1. Let $y = x^2$ and rewrite the equation in terms of $y$.

2. Solve the resulting equation for $y$.

3. Take square roots to solve for $x$.

Try solving on your own before revealing the answer!

Final Answer: The solution set is $\left\{\frac{2\sqrt{3}}{3}, -\frac{\sqrt{5}}{5}, -\frac{2\sqrt{3}}{3}, \frac{\sqrt{5}}{5}\right\}$.

These values are found by solving the quartic and taking square roots.