Algebra 2 quiz answers: 36 practice questions with solutions

These 36 Algebra 2 quiz questions with answers and solutions cover functions, polynomials, quadratics, rational expressions, radicals, and logarithms. Each round focuses on a different skill, with an explanation after every answer so students can review their working.

Round 1: Functions and systems

1. If f(x) = 2x + 3, what is f(4)?
   Answer: 11. Explanation: Substitute 4 for x: f(4) = 2(4) + 3 = 11.
2. Let f(x) = x² and g(x) = x + 2. Find f(g(3)).
   Answer: 25. Explanation: First find g(3) = 5, then evaluate f(5) = 5² = 25.
3. Find the inverse of f(x) = 3x − 6, where x is real.
   Answer: f⁻¹(x) = (x + 6)/3. Explanation: Set y = 3x − 6 and solve for x: x = (y + 6)/3. Swap the variable names to write the inverse.
4. Solve the system x + y = 7 and x − y = 1.
   Answer: (x, y) = (4, 3). Explanation: Add the equations to get 2x = 8, so x = 4. Substitute into x + y = 7 to get y = 3.
5. Find all real intersections of y = x² and y = x + 2.
   Answer: (−1, 1) and (2, 4). Explanation: Set x² = x + 2. Then (x − 2)(x + 1) = 0, so x = 2 or x = −1. Substitute each value into either equation to find y.
6. What is the domain of f(x) = 1/(x − 3) over the real numbers?
   Answer: All real numbers except 3. Explanation: The denominator cannot be zero, so x − 3 ≠ 0.

Round 2: Polynomial operations and factoring

1. Simplify (3x² − 2x + 1) + (x² + 5x − 4).
   Answer: 4x² + 3x − 3. Explanation: Combine the terms with matching powers of x.
2. Expand (x + 2)(x − 3).
   Answer: x² − x − 6. Explanation: Distribute each term: x² − 3x + 2x − 6 = x² − x − 6.
3. Factor x² − 5x + 6.
   Answer: (x − 2)(x − 3). Explanation: The numbers −2 and −3 multiply to 6 and add to −5. Expanding the factors confirms the result.
4. Divide x³ − 8 by x − 2.
   Answer: x² + 2x + 4, with remainder 0. Explanation: The difference-of-cubes identity gives x³ − 8 = (x − 2)(x² + 2x + 4).
5. Use the factor theorem to decide whether x − 3 is a factor of P(x) = x² − 4x + 3.
   Answer: Yes. Explanation: P(3) = 9 − 12 + 3 = 0. Because the remainder is zero, x − 3 is a factor.
6. Find all zeros of P(x) = x³ − 4x.
   Answer: x = −2, 0, and 2. Explanation: Factor P(x) = x(x − 2)(x + 2). Each value makes one factor zero; substituting any of them into the original polynomial confirms it.

Round 3: Quadratic equations and complex numbers

1. Solve x² − 7x + 12 = 0 over the real numbers.
   Answer: x = 3 or x = 4. Explanation: Factor to get (x − 3)(x − 4) = 0. Factoring is convenient here because the integer factors are easy to identify.
2. Solve x² + 6x + 5 = 0 by completing the square.
   Answer: x = −1 or x = −5. Explanation: Rewrite the equation as (x + 3)² = 4. Taking both square roots gives x + 3 = ±2.
3. Use the quadratic formula to solve 2x² + x − 1 = 0.
   Answer: x = 1/2 or x = −1. Explanation: With a = 2, b = 1, and c = −1, the formula gives x = (−1 ± √9)/4. The formula also works when convenient factors are not obvious.
4. Find the vertex of y = x² − 4x + 1.
   Answer: (2, −3). Explanation: Complete the square: y = (x − 2)² − 3. The vertex is at x = 2, y = −3.
5. How many distinct real solutions does x² + 2x + 5 = 0 have?
   Answer: None. Explanation: The discriminant is b² − 4ac = 4 − 20 = −16. A negative discriminant means there are no real roots, though there are two complex roots.
6. Find the complex solutions of x² + 4 = 0, where i² = −1.
   Answer: x = 2i or x = −2i. Explanation: Since x² = −4, its two complex square roots are ±2i.

Round 4: Rational expressions and equations

1. Simplify (x² − 1)/(x − 1), and state the excluded value.
   Answer: x + 1, with x ≠ 1. Explanation: Factor the numerator as (x − 1)(x + 1) and cancel x − 1. The original denominator is zero at x = 1, so that value remains excluded.
2. Simplify 1/x + 1/(x + 1), and state all excluded values.
   Answer: (2x + 1)/(x(x + 1)), with x ≠ 0 and x ≠ −1. Explanation: Use the common denominator x(x + 1): the numerator becomes x + 1 + x = 2x + 1. Neither original denominator can be zero.
3. Solve 2/(x − 1) = 1, stating any excluded value.
   Answer: x = 3; x = 1 is excluded. Explanation: For x ≠ 1, multiply by x − 1 to get 2 = x − 1. Check x = 3 in the original equation: 2/(3 − 1) = 1.
4. Solve 1/x + 1/(x + 2) = 3/4, stating all excluded values.
   Answer: x = 2 or x = −4/3; x = 0 and x = −2 are excluded. Explanation: Multiply by 4x(x + 2) to get 3x² − 2x − 8 = 0, or (3x + 4)(x − 2) = 0. Both solutions meet the restrictions and satisfy the original equation: their left sides are 1/2 + 1/4 and −3/4 + 3/2, respectively.
5. Solve x/(x − 2) = 2/(x − 2), stating any excluded value.
   Answer: No solution; x = 2 is excluded. Explanation: Multiplying by x − 2 suggests x = 2, but that value makes the original denominators zero and cannot be accepted.
6. One worker completes a job alone in 6 hours and another completes it alone in 3 hours. Assuming constant rates and no interference, how long do they take working together?
   Answer: 2 hours. Explanation: Their combined rate is 1/6 + 1/3 = 1/2 job per hour. If t is the time in hours, t > 0 and t/6 + t/3 = 1, so t = 2. Checking gives 2/6 + 2/3 = 1 job.

Round 5: Radicals and rational exponents

1. Evaluate 16^(3/4).
   Answer: 8. Explanation: The principal fourth root of 16 is 2, and 2³ = 8.
2. Simplify √(50x²) for real x.
   Answer: 5|x|√2. Explanation: √50 = 5√2 and √(x²) = |x|. The absolute value is needed because x may be negative.
3. Solve √(x + 5) = 3 over the real numbers.
   Answer: x = 4. Explanation: Squaring gives x + 5 = 9. Check the original equation: √(4 + 5) = 3, and the radicand is nonnegative.
4. Solve √(x + 2) = x over the real numbers.
   Answer: x = 2. Explanation: Since a principal square root is nonnegative, x must be nonnegative. Squaring gives x² − x − 2 = 0, with candidates 2 and −1. Checking the original rejects −1 because √1 ≠ −1; √4 = 2.
5. Simplify x^(1/2) · x^(3/2) for x ≥ 0.
   Answer: x². Explanation: Add the exponents: 1/2 + 3/2 = 2. The stated domain keeps the expressions real.
6. Solve x^(2/3) = 4 over the real numbers.
   Answer: x = −8 or x = 8. Explanation: Let u = ∛x. Then u² = 4, so u = ±2 and x = u³ = ±8. Both values satisfy the original equation.

Round 6: Exponential and logarithmic equations

1. Solve 2^x = 8.
   Answer: x = 3. Explanation: Since 8 = 2³, the exponents are equal.
2. Simplify (3² · 3⁵)/3³.
   Answer: 3⁴ = 81. Explanation: For the common nonzero base 3, add exponents when multiplying and subtract when dividing: 2 + 5 − 3 = 4.
3. Rewrite log₂(32) = 5 in exponential form.
   Answer: 2⁵ = 32. Explanation: A logarithm gives the exponent to which its base must be raised to produce the argument.
4. Solve log₂(x − 1) = 3 and state the domain restriction.
   Answer: x = 9; the domain requires x > 1. Explanation: The logarithm's argument must be positive. Convert to exponential form: x − 1 = 2³ = 8. The result satisfies x > 1.
5. Solve log₁₀(x) + log₁₀(x − 9) = 1 over the real numbers.
   Answer: x = 10; the domain requires x > 9. Explanation: Both arguments must be positive. Combine the logarithms to get x(x − 9) = 10, so (x − 10)(x + 1) = 0. Reject −1 because it is outside the domain; log₁₀(10) + log₁₀(1) = 1.
6. A population starts at 200 and increases by 10% each year. Assuming the rate stays constant, what is the modeled population after 3 years?
   Answer: 266.2, or approximately 266 when rounded to the nearest whole number. Explanation: The exponential model is P(t) = 200(1.10)^t, so P(3) = 200(1.10)³ = 266.2.

How to use and write better Algebra 2 quiz questions

Attempt each question before checking

Cover the answer, show your steps, and then compare your method with the explanation. If your result differs, identify the first step where your work changes and try again.

Choose questions by skill

Select a few questions from one round for targeted practice, or mix rounds for review. For example, pair questions 21 and 23 to check both valid solutions and excluded values.

Adapt questions without changing the skill

When changing an Algebra 2 test question, keep its intended task clear. For example, change the constants in question 21 while retaining a variable in the denominator, then solve the new equation rather than reusing the old answer.

Recheck solutions and restrictions

Substitute proposed solutions into the original equation, not only a transformed version. For grading, specify whether students must show excluded values, identify extraneous solutions, or give exact rather than rounded answers.

If learners need a wider review of prerequisite topics, the math quiz with answers provides mixed practice before more advanced algebra.

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