Use Square and Cube Roots - 8th Grade Math

Between which two consecutive integers does $\sqrt{20}$ lie?
Answer: Between $4$ and $5$ 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 √20, since 4² = 16 and 5² = 25, and 16 < 20 < 25, it lies between 4 and 5. This is correct because the square root preserves inequalities for positive numbers, and 20 is not a perfect square, so √20 is irrational. A common error is misidentifying the bounding squares, like thinking it's between 3 and 4 since 3² = 9 and 4² = 16, but 20 > 16. The strategy is to (1) identify square root, (2) find consecutive integers whose squares bound 20, (3) check perfect squares (memorize up to 12² = 144), (4) note no ± for evaluation, and (5) recognize irrational for non-perfect. Common mistakes include wrong bounding, confusing with cubes, or claiming √2 rational.
Solve for $x$ : $x^3 = 125$ .
Answer: $x=5$ To solve $x^3=125$ , take the cube root of both sides: $x=\sqrt[3]{125}$ . Since $5^3=125$ , the cube root is exactly 5, and a cube root of a positive number has only one real solution, unlike a square root, which has two. Choice C ( $x=\pm5$ ) incorrectly applies the $\pm$ rule from square roots to a cube root. Choice D ( $x=\pm\sqrt{125}$ ) mistakenly uses the square root operation instead of the cube root. Choice B ( $x=25$ ) comes from squaring 5 instead of recognizing that $5^3=125$ .
Evaluate: $\sqrt{81}$
Answer: $9$ This question tests solving $x^2 = p$ and $x^3 = p$ using root notation, evaluating perfect square and cube roots, and recognizing irrational roots. The square root $\sqrt{p}$ solves $x^2 = p$ (for example, if $x^2 = 49$ , then x = $\sqrt{49}$ = 7 or x = - $\sqrt{49}$ = -7, both since $(\pm 7)^2 = 49$ ), while the cube root $\sqrt[3]{p}$ solves $x^3 = p$ (for example, if $x^3 = 64$ , then x = $\sqrt[3]{64}$ = 4 since $4^3 = 64$ , 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 $\sqrt{2} \approx 1.414\ldots$ with non-repeating decimals. Evaluating $\sqrt{81}$ , since $9^2 = 81$ and the principal square root is the positive value, $\sqrt{81} = 9$ . 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 $\pm 9$ , forgetting that $\sqrt{}$ refers to the principal (positive) root. The strategy is to (1) identify the operation as square root, (2) apply $\sqrt{}$ to 81, (3) check if 81 is a perfect square (memorize up to $12^2 = 144$ and $6^3 = 216$ ), (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 $x^2 = p$ , confusing with $\sqrt[3]{}$ , or claiming $\sqrt{2}$ is rational.
Sam solves $5x^3 = 135$ by dividing both sides by 5 to get $x^3 = 27$ , then writes $x = 27^{1/3}$ . Tara solves the same equation and writes $x = \sqrt[3]{27}$ . Who is correct?
Answer: Both are correct; $27^{1/3}$ and $\sqrt[3]{27}$ represent the same value, which is 3 When you encounter equations with exponents like $x^3 = 27$ , 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 $27^{1/3}$ uses fractional exponent notation, where the denominator of the fraction indicates the type of root. Since we have $\frac{1}{3}$ , this means "cube root." Tara's notation $\sqrt[3]{27}$ uses radical notation with the index 3, also indicating cube root. Both expressions equal 3, since $3^3 = 27$ . Looking at the incorrect choices: Choice A wrongly claims that $27^{1/3}$ 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 $27^{1/3} = 9$ (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 $27^{1/3} = \sqrt[3]{27} = 3$ , making both students' approaches mathematically equivalent. Study tip: Remember that fractional exponents and radical notation are interchangeable: $a^{1/n} = \sqrt[n]{a}$ . When you see either form, they represent the same operation and will give you the same answer.
Which is greater?
Answer: $\sqrt{50}$ 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). Comparing √50 and 7, √50 is greater since 7² = 49 and 50 > 49, so √50 > 7. This is correct because the square root function is increasing, and a larger input gives a larger output. A common error is thinking they are equal since 7² is close to 50, or confusing with cubes. To solve, (1) identify square roots, (2) apply √, (3) check squares (7² = 49 < 50, memorize up to 12² = 144), (4) no ± for comparison, (5) recognize √50 irrational and greater than 7. Mistakes include claiming equality, forgetting √50 > √49 = 7, or misidentifying rationality.
Between which two consecutive integers does $\sqrt{50}$ lie?
Answer: Between $7$ and $8$ 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). √50 lies between 7 and 8, since 7² = 49 and 8² = 64, and 50 is between 49 and 64. This is correct because √50 is slightly more than 7 but less than 8. A common error is miscalculating the squares, like thinking it's between 6 and 7 since 6² = 36 and 7² = 49, but 50 > 49. To solve, (1) identify it's a square root, (2) apply √, (3) find perfect squares around 50 (7² = 49, 8² = 64, memorize up to 12² = 144), (4) no ± for evaluation, (5) recognize it's irrational since 50 is non-perfect. Mistakes include wrong interval due to calculation error, confusing with cube roots, or claiming it's rational.
Solve the equation: $x^2 = 49$ . What are all real solutions for $x$ ?
Answer: $x=\pm 7$ 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.
Which equation has a solution that can be expressed exactly using radical notation, but the solution is an irrational number?
Answer: $x^2 = 18$ 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 $\sqrt{18} = \sqrt{9 \times 2} = 3\sqrt{2}$ , and since $\sqrt{2}$ is irrational, the entire expression $3\sqrt{2}$ is irrational. Even though we can write the solution exactly using radical notation, it's not a rational number because $\sqrt{2}$ cannot be expressed as a fraction. Let's examine why the other options don't work. Choice A gives us $\sqrt[3]{8} = 2$ , which is a rational number (an integer). Choice B yields $\sqrt[3]{64} = 4$ , also rational. Choice C produces $\sqrt{25} = 5$ , 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 ( $\pm3\sqrt{2}$ ), but the result is irrational because it contains $\sqrt{2}$ , 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 $\sqrt{2}$ , $\sqrt{3}$ , $\sqrt{5}$ , etc., remain irrational. Look for these "leftover" square roots in your final answer.
Solve the equation: $x^2 = \dfrac{1}{16}$ . What are all real solutions for $x$ ?
Answer: $x=\pm \dfrac{1}{4}$ 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.
Which value is irrational?
Answer: $\sqrt{12}$ 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. Among the options, √12 is irrational because 12 is not a perfect square (simplifies to 2√3, with √3 irrational). This is correct as the others are rational: √36 = 6, ∛8 = 2, √144 = 12, all perfect roots. A common error is misidentifying perfect squares, like thinking √12 is rational or claiming √36 is irrational. The strategy is to (1) identify root type, (2) apply root, (3) check if number is perfect square/cube (memorize up to 12² = 144 and 6³ = 216), (4) note ± not needed for evaluation, and (5) recognize irrational for non-perfect squares. Common mistakes include claiming √2 rational, confusing √ and ∛, or forgetting negatives in equations.
Maria is solving the equation $x^2 = 49$ . She writes $x = \sqrt{49} = 7$ . Her teacher marks this as partially correct. What error did Maria make?
Answer: She should have written $x = \pm\sqrt{49} = \pm 7$ Since $x^2=49$ , both $x=7$ and $x=-7$ satisfy the equation, because $7^2=49$ and $(-7)^2=49$ . Maria's answer only gives the positive root, so the complete solution should be written as $x=\pm\sqrt{49}=\pm7$ . Choice B is wrong because $\sqrt{49}=7$ , not $-7$ ; the principal square root is always the positive one. Choice C is wrong because square roots, not cube roots, are what's needed here. Choice D is wrong because substitution would confirm that 7 works, but it wouldn't reveal that $-7$ also works, which is the actual missing piece.
Solve the equation: $x^3 = 64$ . What is $x$ ?
Answer: $x=4$ 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 x³ = 64, taking the cube root of both sides gives x = ∛64, and since 4³ = 64, x = 4. This is correct because 4 is the only real number whose cube is 64, unlike square roots which have two solutions. A common error is confusing cube roots with square roots and including ±, or miscalculating ∛64 as 8 since √64 = 8. The strategy is to (1) identify the operation as x³, (2) apply the cube root ∛ to both sides, (3) check if 64 is a perfect cube (memorize up to 12² = 144 and 6³ = 216), (4) note no ± for cube roots, and (5) recognize that perfect cubes yield rational roots. Common mistakes include adding a negative solution like for x² = p, using the wrong root symbol like √ instead of ∛, or claiming √2 is rational.
Which expression is an irrational number?
Answer: $\sqrt{12}$ This question tests solving $x^2 = p$ and $x^3 = p$ using root notation, evaluating perfect square and cube roots, and recognizing irrational roots. The square root $√p$ solves $x^2 = p$ (if $x^2 = 49$ , then $x = √49 = 7$ or $x = -√49 = -7$ , both since $(±7)^2 = 49$ ), while the cube root $∛p$ solves $x^3 = p$ (if $x^3 = 64$ , then $x = ∛64 = 4$ since $4^3 = 64$ , 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= $4^2$ , 144= $12^2$ , 64= $4^3$ ; memorize up to 12 $^2$ =144, 6 $^3$ =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.
Which statement is true?
Answer: $\sqrt{2}$ is irrational. 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 true statement is that √2 is irrational, as 2 is not a perfect square. This is correct because √2 cannot be expressed as a fraction and has a non-repeating decimal, unlike √9 = 3 or √4 = 2 which are rational. A common error is misidentifying perfect squares like claiming √3 is rational or √9 is irrational. To solve, (1) identify square roots, (2) apply √, (3) check if the number is a perfect square (memorize up to 12² = 144), (4) include ± only for equations, (5) recognize irrational if non-perfect gives non-repeating decimal. Mistakes include claiming √2 rational, confusing rational and irrational, or forgetting which are perfect squares.
The equation $2x^3 - 54 = 0$ can be rewritten as $x^3 = 27$ . A student claims that since $\sqrt{27}$ is irrational, the solution $x = \sqrt[3]{27}$ must also be irrational. What is wrong with this reasoning?
Answer: The student confused square roots with cube roots; $\sqrt[3]{27} = 3$ is rational The student mixed up square roots and cube roots. It's true that $\sqrt{27}=\sqrt{9\times3}=3\sqrt{3}$ is irrational, but $\sqrt[3]{27}=\sqrt[3]{3^3}=3$ is rational, because 27 is a perfect cube, even though it isn't a perfect square. Choice B is wrong because $\sqrt[3]{27}$ is rational, not irrational. Choice C is wrong because $x^3=27$ is set up correctly from $2x^3-54=0$ ; there's no algebra error. Choice D is wrong because checking whether 27 is a perfect square is the wrong test for a cube root; what matters is whether 27 is a perfect cube, which it is.
Solve for $x$ : $x^2 = 49$ .
Answer: $x=\pm 7$ 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.
The expression $\sqrt{x^2}$ is sometimes written as $|x|$ rather than just $x$ . For which value of $x$ does this distinction matter when solving $x^2 = 16$ ?
Answer: $x = -4$ , because $\sqrt{(-4)^2} = \sqrt{16} = 4 = |-4|$ , not $-4$ When working with square roots and absolute values, you need to understand a crucial distinction: $\sqrt{x^2}$ always gives you the positive result, which is why it equals $|x|$ , not necessarily $x$ itself. Let's solve $x^2 = 16$ to see where this matters. Taking the square root of both sides: $x = \pm\sqrt{16} = \pm 4$ . So our solutions are $x = 4$ and $x = -4$ . Now let's check what happens with $\sqrt{x^2}$ for each solution. When $x = 4$ : $\sqrt{4^2} = \sqrt{16} = 4$ , and $|4| = 4$ . Both give us 4, so the distinction doesn't matter here. When $x = -4$ : $\sqrt{(-4)^2} = \sqrt{16} = 4$ , and $|-4| = 4$ . Here's the key insight—even though $x = -4$ , both $\sqrt{x^2}$ and $|x|$ give us positive 4, not negative 4. This shows why we write $\sqrt{x^2} = |x|$ 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 $\sqrt{0^2} = 0$ is perfectly defined. Choice D has multiple errors: 8 isn't even a solution to $x^2 = 16$ , and $|8| = 8$ , not $-8$ . Study tip: Remember that $\sqrt{x^2} = |x|$ always. The square root operation gives you the positive value, which matters most when dealing with negative inputs.
A scientist needs to find the value of $n$ where $n^3 + n^2 = 72$ . She notices that if $n = 3$ , then $n^3 = 27$ and $n^2 = 9$ . What should she conclude?
Answer: Since $27 + 9 = 36 \neq 72$ , she needs $n > 3$ and should test $n = 4$ next When solving polynomial equations like $n^3 + n^2 = 72$ , 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 $n = 3$ : $n^3 = 27$ and $n^2 = 9$ , so $n^3 + n^2 = 27 + 9 = 36$ . Since $36 < 72$ , she needs a larger value of $n$ . This makes sense because both $n^3$ and $n^2$ 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 $36 = \frac{72}{2}$ , you can simply double $n$ from 3 to 6. This linear thinking doesn't work with polynomial equations because $n^3$ and $n^2$ don't scale linearly. If $n = 6$ , then $6^3 + 6^2 = 216 + 36 = 252$ , which is way too large. Choice D suggests jumping directly to $\sqrt[3]{72}$ , but this would only work if the equation were $n^3 = 72$ , not $n^3 + n^2 = 72$ . The $n^2$ 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.
Evaluate: $\sqrt[3]{125}$
Answer: $5$ This question tests solving $x^2 = p$ and $x^3 = p$ using root notation, evaluating perfect square and cube roots, and recognizing irrational roots. The square root $\sqrt{p}$ solves $x^2 = p$ (for example, if $x^2 = 49$ , then $x = \sqrt{49} = 7$ or $x = -\sqrt{49} = -7$ , both since $(±7)^2 = 49$ ), while the cube root $\sqrt[3]{p}$ solves $x^3 = p$ (for example, if $x^3 = 64$ , then $x = \sqrt[3]{64} = 4$ since $4^3 = 64$ , 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 $\sqrt{2} \approx 1.414\ldots$ with non-repeating decimals. Evaluating $\sqrt[3]{125}$ , since $5^3 = 125$ , $\sqrt[3]{125} = 5$ . 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 $\pm 5$ , or miscalculating as 25 since $5^2 = 25$ . The strategy is to (1) identify the operation as cube root, (2) apply $\sqrt[3]{}$ to 125, (3) check if 125 is a perfect cube (memorize up to $12^2 = 144$ and $6^3 = 216$ ), (4) note no $\pm$ for cube roots, and (5) recognize rational for perfect cubes. Common mistakes include adding $\pm$ like for $x^2 = p$ , using $\sqrt{}$ instead of $\sqrt[3]{}$ , or claiming $\sqrt{2}$ is rational.
A science club builds a cube-shaped container with volume $216\text{ cm}^3$ . What is the side length of the cube?
Answer: $6\text{ cm}$ because $\sqrt[3]{216}=6$ This question tests solving $x^2 = p$ and $x^3 = p$ using root notation, evaluating perfect square and cube roots, and recognizing irrational roots. The square root $\sqrt{p}$ solves $x^2 = p$ (for example, if $x^2 = 49$ , then $x = \sqrt{49} = 7$ or $x = -\sqrt{49} = -7$ , both since $(\pm 7)^2 = 49$ ), while the cube root $\sqrt[3]{p}$ solves $x^3 = p$ (for example, if $x^3 = 64$ , then $x = \sqrt[3]{64} = 4$ since $4^3 = 64$ , 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 $\sqrt{2} \approx 1.414\ldots$ with non-repeating decimals. For a cube with volume 216 cm³, the side length is $\sqrt[3]{216}$ , and since $6^3 = 216$ , 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 $\sqrt[3]{216}$ as 8 (since $8^3 = 512$ ), or adding ± for length. The strategy is to (1) identify $x^3$ for volume, (2) apply $\sqrt[3]{}$ to 216, (3) check if perfect cube (memorize up to $6^3 = 216$ ), (4) note no ± for lengths, and (5) recognize rational for perfect cubes. Common mistakes include using $\sqrt{}$ instead of $\sqrt[3]{}$ , forgetting negatives in equations, or claiming $\sqrt{2}$ rational.

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