Practice Use Square And Cube Roots in 8th Grade Math with focused quiz questions that help you check what you know, review explanations, and build confidence with test-style prompts.
What this quiz covers
This quiz focuses on Use Square And Cube Roots, giving you a quick way to practice the rules, question types, and explanations that matter most for 8th Grade Math.
How to use this quiz
Try each quiz question before looking at the correct answer. Use the explanations to review missed ideas, then come back to similar questions until the pattern feels familiar.
Question 1
The expression x2 is sometimes written as ∣x∣ rather than just x. For which value of x does this distinction matter when solving x2=16?
x=4, because 42
Explanation: When working with square roots and absolute values, you need to understand a crucial distinction: x2 always gives you the positive result, which is why it equals , not necessarily itself.
Question 2
A cube has a volume of 216 cubic inches. To find the side length, Alex sets up the equation s3=216 and concludes that s=3. What is the next step to find the exact value?
Question 3
Maria is solving the equation x2=49. She writes x=49. Her teacher marks this as partially correct. What error did Maria make?
Question 4
A storage container is designed so that its square base has area A=144 square feet and its height equals the side length of the base. What is the volume of the container?
144 cubic feet, found by calculating V=A2 divided by 144
Question 5
The equation 2x3−54=0 can be rewritten as x3=. A student claims that since is irrational, the solution must also be irrational. What is wrong with this reasoning?
Question 6
A square garden has the same area as a circle with radius 6 feet. If the side length of the square is s, which equation correctly represents this situation?
s2=6π, so feet exactly
Question 7
Which equation has a solution that can be expressed exactly using radical notation, but the solution is an irrational number?
x3=8, with solution x=
Question 8
A scientist needs to find the value of n where n3+n2=72. She notices that if , then and . What should she conclude?
Question 9
Sam solves 5x3=135 by dividing both sides by 5 to get x3=27, then writes . Tara solves the same equation and writes . Who is correct?
Question 10
Solve the equation: x2=161. What are all real solutions for x?
Question 11
A square has area 50 cm2. The side length is 50 cm. Which comparison is correct?
Question 12
A square has area 20 cm2. What is the exact side length of the square?
20 cm
Question 13
A science club builds a cube-shaped container with volume 216 cm3. What is the side length of the cube?
6 cm because
Question 14
Which expression is an irrational number?
12
Question 15
Evaluate: 3125
Question 16
Evaluate: 81
±9
Question 17
Solve the equation: x2=49. What are all real solutions for x?
x=
Question 18
Evaluate 121.
±11
Question 19
Evaluate 3216.
Question 20
Solve for x: x2=49.
x=
=
16=
4=
∣4∣
x=−4, because (−4)2=16=4=∣−4∣, not −4
x=0, because 02 is undefined while ∣0∣=0
x=8, because 82=8 but ∣8∣=−8 for this equation
∣x∣
x
Let's solve x2=16 to see where this matters. Taking the square root of both sides: x=±16=±4. So our solutions are x=4 and x=−4.
Now let's check what happens with x2 for each solution. When x=4: 42=16, and ∣4∣=4. Both give us 4, so the distinction doesn't matter here.
When x=−4: (−4)2=16=4, and ∣−4∣=4. Here's the key insight—even though x=−4, both x2 and ∣x∣ give us positive 4, not negative 4. This shows why we write x2=∣x∣ rather than just x. The distinction matters for negative values because the square root symbol always returns the positive result.
Choice A is wrong because ∣4∣=4, so there's no difference. Choice C is incorrect since 02=0 is perfectly defined. Choice D has multiple errors: 8 isn't even a solution to x2=16, and ∣8∣=8, not −8.
Study tip: Remember that x2=∣x∣ always. The square root operation gives you the positive value, which matters most when dealing with negative inputs.
216
Calculate 3216=363=6 inches exactly
Use a calculator to approximate 3216≈6.00 inches
Rewrite as s=216= inches
Factor out perfect squares: 3216= inches
Explanation: Since 216=63, we can evaluate 3216=363=6 exactly. This gives the exact side length as 6 inches. Choice B uses approximation when an exact answer is available. Choice C incorrectly uses square root instead of cube root. Choice D incorrectly factors and doesn't recognize that 216 is a perfect cube.
=
7
She should have written x=±49=±7
She calculated 49 incorrectly as 7 instead of -7
She should have cubed both sides instead of taking square roots
She forgot to check her answer by substituting back into the equation
Explanation: When solving x2=49, both positive and negative values of x satisfy the equation since 72=49 and (−7)2=49. The complete solution is x=±49=±7. Choice B is incorrect because 49=7 (principal square root is positive). Choice C is wrong because square roots, not cube roots, are needed. Choice D is incorrect because substitution would verify the answer but doesn't address the missing negative solution.
=
1442
1440 cubic feet, found by calculating V=A×h=144×10
432 cubic feet, found by calculating V=A×A=144×3
1728 cubic feet, found by calculating s=144=12 and V=s3=123
Explanation: When you encounter a volume problem involving a container with specific dimensions, start by identifying what you know and what relationships exist between the measurements. Here, you have a square base with area 144 square feet, and the height equals the side length of the base.To find the volume, you first need the side length of the square base. Since the area of a square is s2, you can find the side length: s=144=12 feet. The problem states that the height equals this side length, so h=12 feet as well. The volume of a rectangular container is V=length×width×height, which for a square base becomes V=s2×h=s3=12 cubic feet.Answer choice A incorrectly squares the area and then divides by 144, which has no geometric meaning for volume calculations. Choice B assumes the height is 10 feet without justification—it ignores the given relationship that height equals the side length. Choice C calculates 144×3, but 144=12, not 3, showing a square root error. Choice D correctly identifies that and calculates cubic feet.Remember: when working with volume problems, always establish all dimensions clearly before applying the volume formula. Pay special attention to relationships between dimensions—don't assume values that aren't given or derivable from the problem statement.
27
27
x=327
The student confused square roots with cube roots; 327=3 is rational
The student is correct; both 27 and 327 are irrational numbers
The equation should be x3=54, not x3=27, leading to different conclusions
The student incorrectly assumed that 27 is irrational when it equals 3
Explanation: The student incorrectly applied properties of square roots to cube roots. While 27=9×3=33 is indeed irrational, 327=3 is rational because 27 is a perfect cube. Choice B is wrong because 327 is rational. Choice C incorrectly suggests an algebraic error. Choice D is wrong because 27=3; it equals 33.
s=
6π
s2=12π, so s=12π=23π feet
s2=36π, so s=36π=6π feet
s=6π, so the side length is 6π feet directly
Explanation: When you see problems asking you to set two areas equal to each other, you need to write the area formulas for both shapes and create an equation.First, find the area of the circle with radius 6 feet. The area formula for a circle is A=πr2, so the area is π⋅62=36π square feet. Next, the area of a square with side length s is s2. Since the areas are equal, you can write: s2=36π. Solving for s, you get s=36π=6π feet. This matches answer choice C.Let's examine why the other answers are wrong. Choice A uses s2=6π, which incorrectly uses just the radius (6) times π, forgetting to square the radius in the circle area formula. Choice B gives s2=, which appears to use instead of —this might come from confusing area with circumference formulas. Choice D sets directly, which completely skips the area calculation and incorrectly equates the side length to the circle's circumference rather than working with areas.Remember this pattern: when setting areas equal, always write out both area formulas completely before creating your equation. Double-check that you're using area formulas (πr2 for circles, s2 for squares) rather than perimeter formulas, since mixing these up is a common mistake on geometry problems.
3
8
=
2
x3=64, with solution x=364=4
x2=25, with solution x=±25=±5
x2=18, with solution x=±18=±32
Explanation: This question tests your understanding of rational versus irrational numbers and how they appear when expressed in radical form. When you see radicals, you need to determine whether they simplify to rational numbers (like integers or fractions) or remain irrational.The key is recognizing that 18=9×2=32, and since 2 is irrational, the entire expression 32 is irrational. Even though we can write the solution exactly using radical notation, it's not a rational number because 2 cannot be expressed as a fraction.Let's examine why the other options don't work. Choice A gives us 38=2, which is a rational number (an integer). Choice B yields , also rational. Choice C produces , again rational. In all these cases, the radicals simplify completely to whole numbers, making the solutions rational despite being written in radical form initially.Only choice D satisfies both conditions: the solution can be expressed exactly using radicals (±32), but the result is irrational because it contains 2, which cannot be simplified further to a rational number.Study tip: When evaluating radicals, always check if they simplify to rational numbers. Perfect squares and perfect cubes under radicals will give rational results, but numbers like 2, 3, , etc., remain irrational. Look for these "leftover" square roots in your final answer.
n=3
n3=27
n2=9
Since 327+9=3+3=6=72, the equation has no solution
Since 27+9=36=72, she needs n>3 and should test next
Since 27+9=36=272, the solution is
Since 27+9=36<72, she should try n=3 directly
Explanation: When solving polynomial equations like n3+n2=72, you often need to test values systematically since these can't always be solved with simple algebra at your level.The scientist correctly calculated that when n=3: n3=27 and n2=9, so n3+n2=27+9=36. Since 36<72, she needs a larger value of n. This makes sense because both n3 and n2 increase as n increases, so their sum will also increase. The logical next step is testing n=4, making choice B correct.Choice A incorrectly takes cube roots and square roots of the results (27 and 9) instead of using the original values. This completely changes the equation and isn't relevant to the problem.Choice C uses flawed reasoning by assuming that since 36=272, you can simply double n from 3 to 6. This linear thinking doesn't work with polynomial equations because n and don't scale linearly. If , then , which is way too large.Choice D suggests jumping directly to 372, but this would only work if the equation were n, not . The term makes this approach incorrect.Study tip: When solving polynomial equations by testing values, work systematically with integers first. If your test value gives a result that's too small, try the next larger integer before attempting more complex approaches.
x=271/3
x=327
Only Tara is correct; 327=3 while 271/3 represents repeated multiplication
Only Sam is correct; 271/3=9 while 327 is undefined for positive numbers
Both are correct; 271/3 and 327 represent the same value, which is 3
Neither is correct; the equation x3=27 should be solved using logarithms, not radicals
Explanation: When you encounter equations with exponents like x3=27, there are two equivalent ways to express the solution using different notation for the same mathematical operation.Both Sam and Tara are using correct notation to find the cube root of 27. The expression 271/3 uses fractional exponent notation, where the denominator of the fraction indicates the type of root. Since we have 31, this means "cube root." Tara's notation 327 uses radical notation with the index 3, also indicating cube root. Both expressions equal 3, since 33=27.Looking at the incorrect choices: Choice A wrongly claims that 271/3 represents repeated multiplication rather than a root operation. This shows a misunderstanding of fractional exponents. Choice B contains two major errors: it incorrectly states that 271/3=9 (when it actually equals 3), and falsely claims cube roots are undefined for positive numbers (cube roots exist for all real numbers). Choice D incorrectly suggests logarithms are needed when simple root operations work perfectly for this type of equation.The key insight is that 271/3=327, making both students' approaches mathematically equivalent.Study tip: Remember that fractional exponents and radical notation are interchangeable: a1/n=na. When you see either form, they represent the same operation and will give you the same answer.
x=−41
x=±41
x=41
x=±161
Explanation: This question tests solving x² = p and x³ = p using root notation, evaluating perfect square and cube roots, and recognizing irrational roots. The square root √p solves x² = p (for example, if x² = 49, then x = √49 = 7 or x = -√49 = -7, both since (±7)² = 49), while the cube root ∛p solves x³ = p (for example, if x³ = 64, then x = ∛64 = 4 since 4³ = 64, with only one real solution); perfect squares like 4, 9, 16, 25, 36, 49, 64 have rational square roots, while non-perfect ones like 2, 3, 5, 6, 7, 8 have irrational roots such as √2 ≈ 1.414... with non-repeating decimals. For x² = 1/16, taking square roots gives x = ±√(1/16), and since √(1/16) = 1/4 (because (1/4)² = 1/16), x = ±1/4. This is correct because 1/16 is a perfect square (fractionally), and both positive and negative satisfy the equation. A common error is choosing ±1/16, confusing the root with squaring again, or forgetting the negative solution. The strategy is to (1) identify x² operation, (2) apply √ to both sides, (3) check if 1/16 is a perfect square (1/4² = 1/16), (4) include ± for solutions, and (5) recognize rational since perfect. Common mistakes include forgetting negative for x² = p, using ∛ wrongly, or claiming √2 rational.
50>8 because 64=8
50<7 because 50<49
50=7 because 50 is close to
50>7 because and
Explanation: This question tests solving x2=p and x3=p using root notation, evaluating perfect square and cube roots, and recognizing irrational roots. The square root p solves x2=p (for example, if x2=49, then x=49=7 or x=−49=−7, both since (±7)2=49), while the cube root 3p solves x3=p (for example, if x3=64, then x=364=4 since 43=64, with only one real solution); perfect squares like 4, 9, 16, 25, 36, 49, 64 have rational square roots, while non-perfect ones like 2, 3, 5, 6, 7, 8 have irrational roots such as 2≈1.414… with non-repeating decimals. For a square with area 50 cm2, the side is 50 cm, and since 49=7 and 50>49, 50>7. This is correct because the square root function is increasing, so a larger input gives a larger output, and 50 is not a perfect square, making 50 irrational. A common error is thinking 50<7 because of miscomparing 50 and 49, or approximating incorrectly like saying it's greater than 8. The strategy is to (1) identify the square root context, (2) apply to 50, (3) compare to nearby perfect squares (72=49, 82=64), (4) note no ± for length, and (5) recognize irrational for non-perfect. Common mistakes include forgetting irrationality, using wrong comparisons, or claiming 2 rational.
20 cm
±20 cm
10 cm
Explanation: This question tests solving x² = p and x³ = p using root notation, evaluating perfect square and cube roots, and recognizing irrational roots. The square root √p solves x² = p (if x² = 49, then x = √49 = 7 or x = -√49 = -7, both since (±7)² = 49), while the cube root ∛p solves x³ = p (if x³ = 64, then x = ∛64 = 4 since 4³ = 64, only one solution); perfect squares like 4, 9, 16, 25, 36, 49, 64 have rational square roots, while non-perfect ones like 2, 3, 5, 6, 7, 8 have irrational roots (√2 ≈ 1.414... non-repeating). For a square with area 20 cm², the side length is √20 cm, the positive square root. This is correct because side length is a physical measurement that must be positive, and area = side², so side = √area. A common error is including the negative ±√20 or confusing with √10. To solve, (1) identify it's x² = 20 for side x, (2) apply √, (3) check 20 not perfect (between 4²=16 and 5²=25), (4) use positive for length, (5) recognize irrational since non-perfect. Mistakes include adding negative for length, using wrong value like 20, or claiming rational.
3
216
=
6
±6 cm
36 cm because 216=36
8 cm because 3216=8
Explanation: This question tests solving x2=p and x3=p using root notation, evaluating perfect square and cube roots, and recognizing irrational roots. The square root p solves x2=p (for example, if x2=49, then x=49=7 or x=−49=−7, both since (±7)2=49), while the cube root 3p solves x3=p (for example, if x3=64, then x=364=4 since 43=64, with only one real solution); perfect squares like 4, 9, 16, 25, 36, 49, 64 have rational square roots, while non-perfect ones like 2, 3, 5, 6, 7, 8 have irrational roots such as 2≈1.414… with non-repeating decimals. For a cube with volume 216 cm³, the side length is 3216, and since 63=216, it is 6 cm. This is correct because volume of a cube is side³, so cube root gives the side, and 216 is a perfect cube. A common error is confusing with square root and choosing 36 or miscalculating 3216 as 8 (since 83=512), or adding ± for length. The strategy is to (1) identify x3 for volume, (2) apply 3 to 216, (3) check if perfect cube (memorize up to 63=216), (4) note no ± for lengths, and (5) recognize rational for perfect cubes. Common mistakes include using instead of 3, forgetting negatives in equations, or claiming 2 rational.
364
16
144
Explanation: This question tests solving x2=p and x3=p using root notation, evaluating perfect square and cube roots, and recognizing irrational roots. The square root √p solves x2=p (if x2=49, then x=√49=7 or x=−√49=−7, both since (±7)2=49), while the cube root ∛p solves x3=p (if x3=64, then x=∛64=4 since 43=64, only one solution); perfect squares like 4, 9, 16, 25, 36, 49, 64 have rational square roots, while non-perfect ones like 2, 3, 5, 6, 7, 8 have irrational roots (√2≈1.414... non-repeating). The irrational expression is √12, since 12 is not a perfect square. This is correct because √12 simplifies to 2√3, which is irrational, unlike √16=4, √144=12, or ∛64=4 which are rational integers. A common error is misidentifying perfect squares, like thinking √12 is rational or choosing ∛64 as irrational. To solve, (1) identify roots, (2) apply √ or ∛, (3) check perfect (12 not square, but 16=42, 144=122, 64=43; memorize up to 122=144, 63=216), (4) no ± for evaluation, (5) recognize irrational if non-perfect square. Mistakes include claiming √2 rational equivalent for others, confusing square and cube roots, or forgetting non-perfect means irrational.
±5
5
25
125
Explanation: This question tests solving x2=p and x3=p using root notation, evaluating perfect square and cube roots, and recognizing irrational roots. The square root p solves x2=p (for example, if x2=49, then x=49=7 or x=−49=−7, both since (±7)2=49), while the cube root 3p solves x3=p (for example, if x3=64, then x=364=4 since 43=64, with only one real solution); perfect squares like 4, 9, 16, 25, 36, 49, 64 have rational square roots, while non-perfect ones like 2, 3, 5, 6, 7, 8 have irrational roots such as 2≈1.414… with non-repeating decimals. Evaluating 3125, since 53=125, 3125=5. This is correct because 125 is a perfect cube, and cube roots have a single real value. A common error is confusing with square roots and choosing ±5, or miscalculating as 25 since 52=25. The strategy is to (1) identify the operation as cube root, (2) apply 3 to 125, (3) check if 125 is a perfect cube (memorize up to 122=144 and 63=216), (4) note no ± for cube roots, and (5) recognize rational for perfect cubes. Common mistakes include adding ± like for x2=p, using instead of 3, or claiming 2 is rational.
9
−9
8
Explanation: This question tests solving x2=p and x3=p using root notation, evaluating perfect square and cube roots, and recognizing irrational roots. The square root p solves x2=p (for example, if x2=49, then x = 49 = 7 or x = -49 = -7, both since (±7)2=49), while the cube root 3p solves x3=p (for example, if x3=64, then x = 364 = 4 since 43=64, with only one real solution); perfect squares like 4, 9, 16, 25, 36, 49, 64 have rational square roots, while non-perfect ones like 2, 3, 5, 6, 7, 8 have irrational roots such as 2≈1.414… with non-repeating decimals. Evaluating 81, since 92=81 and the principal square root is the positive value, 81=9. This is correct because 81 is a perfect square, and the square root symbol denotes the non-negative root. A common error is including the negative root like -9 or ±9, forgetting that refers to the principal (positive) root. The strategy is to (1) identify the operation as square root, (2) apply to 81, (3) check if 81 is a perfect square (memorize up to 122=144 and 63=216), (4) use only the positive for principal square root, and (5) recognize rational for perfect squares. Common mistakes include adding a negative like for solving x2=p, confusing with 3, or claiming 2 is rational.
7
x=±49 only means x=49
x=±7
x=−7
Explanation: This question tests solving x² = p and x³ = p using root notation, evaluating perfect square and cube roots, and recognizing irrational roots. The square root √p solves x² = p (for example, if x² = 49, then x = √49 = 7 or x = -√49 = -7, both since (±7)² = 49), while the cube root ∛p solves x³ = p (for example, if x³ = 64, then x = ∛64 = 4 since 4³ = 64, with only one real solution); perfect squares like 4, 9, 16, 25, 36, 49, 64 have rational square roots, while non-perfect ones like 2, 3, 5, 6, 7, 8 have irrational roots such as √2 ≈ 1.414... with non-repeating decimals. For x² = 49, taking the square root of both sides gives x = ±√49, and since 49 is 7², √49 = 7, so x = ±7. This is correct because both 7² = 49 and (-7)² = 49 satisfy the equation, providing all real solutions. A common error is forgetting the negative solution and choosing only x = 7, or confusing it with cube roots which have only one real root. The strategy is to (1) identify the operation as x², (2) apply the square root √ to both sides, (3) check if 49 is a perfect square (memorize up to 12² = 144 and 6³ = 216), (4) include ± for square roots, and (5) recognize that perfect squares yield rational roots. Common mistakes include forgetting the negative solution for x² = p, using the wrong root symbol like ∛ instead of √, or claiming √2 is rational.
−11
11
12
Explanation: This question tests solving x2=p and x3=p using root notation, evaluating perfect square and cube roots, and recognizing irrational roots. The square root p solves x2=p (if x2=49, then x = 49 = 7 or x = -49 = -7, both since (±7)2 = 49), while the cube root 3p solves x3=p (if x3=64, then x = 364 = 4 since 43 = 64, only one solution); perfect squares like 4, 9, 16, 25, 36, 49, 64 have rational square roots, while non-perfect ones like 2, 3, 5, 6, 7, 8 have irrational roots (2≈1.414… non-repeating). Evaluating 121 gives 11, since 112 = 121 and the principal square root is the positive value. This is correct because the square root symbol denotes the non-negative root, not the negative or both. A common error is including the negative or both, like ±11, confusing evaluation with solving equations. To solve, (1) identify it's a square root, (2) apply , (3) check if 121 is a perfect square (112 = 121, memorize up to 122 = 144), (4) no ± for evaluation of , (5) recognize it's rational since it's perfect. Mistakes include adding negative solutions when just evaluating p, confusing with cube roots, or claiming it's irrational.
8
6
36
±6
Explanation: This question tests solving x2=p and x3=p using root notation, evaluating perfect square and cube roots, and recognizing irrational roots. The square root p solves x2=p (if x2=49, then x=49=7 or x=−49=−7, both since (±7)2=49), while the cube root 3p solves x3=p (if x3=64, then x=364=4 since 43=64, only one solution); perfect squares like 4, 9, 16, 25, 36, 49, 64 have rational square roots, while non-perfect ones like 2, 3, 5, 6, 7, 8 have irrational roots (2≈1.414… non-repeating). Evaluating 3216 gives 6, since 63=216. This is correct because cube roots have a single real value, and 216 is a perfect cube. A common error is confusing with square roots and adding ±6 or choosing a square like 36. To solve, (1) identify it's a cube root, (2) apply 3, (3) check if 216 is a perfect cube (63=216, memorize up to 63=216), (4) no ± for cube roots, (5) recognize it's rational since it's perfect. Mistakes include adding ± like for square roots, using the wrong root symbol, or miscalculating as 8.
±7
x=7
x=±49=±6
x=−7
Explanation: This question tests solving x² = p and x³ = p using root notation, evaluating perfect square and cube roots, and recognizing irrational roots. The square root √p solves x² = p (if x² = 49, then x = √49 = 7 or x = -√49 = -7, both since (±7)² = 49), while the cube root ∛p solves x³ = p (if x³ = 64, then x = ∛64 = 4 since 4³ = 64, only one solution); perfect squares like 4, 9, 16, 25, 36, 49, 64 have rational square roots, while non-perfect ones like 2, 3, 5, 6, 7, 8 have irrational roots (√2 ≈ 1.414... non-repeating). For x² = 49, taking the square root gives x = ±√49, and since 49 is a perfect square (7² = 49), x = ±7. This is correct because both 7² and (-7)² equal 49, so both solutions apply. A common error is forgetting the negative and choosing only x = 7, or miscalculating √49 as 6. To solve, (1) identify x², (2) apply √, (3) check perfect square (7² = 49, memorize up to 12² = 144), (4) include ±, (5) recognize rational. Mistakes include omitting negative, using wrong root, or claiming irrational.